Chapter test · nothing is marked until you submit

Quadratic Functions and Equations: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 138 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. Three costumes, and what each one hands over . 12 points. Question 1 of 10.

The quadratic function $f(x) = x^2 - 10x + 21$ arrives in standard form. Standard form is only one of the costumes it can wear, and the parts below are about which question each costume answers without any work at all.
1. Part A.
  Put $f$ into factored form, and give the axis of symmetry of its graph.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  State the vertex of the graph, its range, its two $x$ -intercepts and its $y$ -intercept.
  
  Carry your own answer forward Read the features off your own forms from the previous part, and say which form each reading came from. What earns credit is taking a feature from the form that displays it, rather than matching one particular list.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  The three forms of a quadratic each answer one of the questions above at sight, and each conceals another. For every one of the three, name a fact it hands over for nothing and a fact it hides.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer
Part A

$f(x) = (x - 3)(x - 7)$ , and the axis of symmetry is $x = 5$ .
- $(x - 7)(x - 3)$ is the same product with its factors written the other way round; the axis is a vertical LINE, so $x = 5$ rather than the bare number $5$
Part B

Vertex $(5, -4)$ , range $[-4, \infty)$ , $x$ -intercepts $3$ and $7$ , $y$ -intercept $21$ .
- the range may be written $y \ge -4$ ; a range of all real numbers belongs to no quadratic, since the outputs are bounded on one side
Part C

Standard form gives the $y$ -intercept $c$ and the end behavior, and hides the vertex. Vertex form gives the vertex, the axis, the extreme value and the range, and hides the crossings. Factored form gives the crossings and the sign of $f$ on each piece of the line, and hides the height of the vertex.
Worked solution

Part A

Two numbers multiplying to $21$ and adding to $-10$ are $-3$ and $-7$ .

$x^2 - 10x + 21 = (x - 3)(x - 7)$

The factored form displays the two zeros, $3$ and $7$ , and two inputs sharing an output are mirror partners, so the axis stands midway between them at $x = \tfrac{3 + 7}{2} = 5$ . Reading the axis off the standard form agrees, since $-\tfrac{b}{2a} = -\tfrac{-10}{2} = 5$ .

Part B

The vertex sits on the axis, so evaluate there.

$f(5) = 25 - 50 + 21 = -4 \quad \Longrightarrow \quad \text{vertex } (5, -4)$

The leading coefficient is $1 > 0$ , so the parabola opens upward, the vertex is its lowest point and the range is $[-4, \infty)$ . The factored form gives the crossings at $3$ and $7$ , and standard form gives the $y$ -intercept for nothing, since $f(0) = c = 21$ .

Part C

Standard form $ax^2 + bx + c$ . Substituting $x = 0$ kills every term carrying an $x$ , so $f(0) = c$ is the $y$ -intercept at no cost, and the sign of $a$ settles the end behavior. What it hides is the middle of the picture: the turning point is nowhere on display.

Vertex form $a(x - h)^2 + k$ . The parent $y = x^2$ has been shifted to put its turning point at $(h, k)$ , so the vertex, the axis $x = h$ , the extreme value $k$ and the range are all free. What it hides is the exact $x$ -coordinates of the crossings, when crossings exist.

Factored form $a(x - r_1)(x - r_2)$ . Substituting $r_1$ or $r_2$ kills a factor, so the crossings are on display, and each factor keeps a fixed sign on either side of its own root, which fixes the sign of $f$ on each piece of the line. What it hides is how high or low the vertex sits.

$f(x) = x^2 - 10x + 21 = (x - 5)^2 - 4 = (x - 3)(x - 7)$

One object, three costumes: choosing a form is choosing which question you get to answer without working for it.

In one line

$f(x) = x^2 - 10x + 21 = (x - 3)(x - 7)$ , with axis $x = 5$ , vertex $(5, -4)$ , range $[-4, \infty)$ , crossings at $3$ and $7$ and $y$ -intercept $21$ . Standard form hands over the $y$ -intercept and the end behavior, vertex form the vertex and everything that follows from it, and factored form the crossings and the sign of $f$ on each piece of the line, while each of the three hides something one of the others puts on show.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Produces two linear factors whose expansion returns the original expression. . Worth 2 points.

Locates the axis as the midpoint of the two zeros, or as $-\tfrac{b}{2a}$ , and reports it as a vertical line. . Worth 2 points.

Part B 4 points

Takes each feature from a form that displays it, evaluating on the axis for the vertex rather than guessing its height. . Worth 2 points.

Reports the range as an interval closed at the correct end, the end decided by the sign of the leading coefficient. . Worth 2 points.

Part C 4 points

Names, for each of the three forms, a fact it hands over at sight, and grounds that in the structure of the form rather than in a remembered list. . Worth 3 points. needs an explanation, not just an answer

Names, for each of the three forms, something it conceals. . Worth 1 point.
2. Two equations that differ only in one number . 15 points. Question 2 of 10.

Two equations arrive side by side:

$x^2 + 5x = 24 \qquad \text{and} \qquad x^2 + 5x = 20.$

The left sides are identical. Only the number on the right has moved.
1. Part A.
  Solve the first equation.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Solve the second equation, giving exact values.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  The two equations have the same left side, and a search for a pair of integers finished the first and cannot finish the second. Explain what decides whether such a search can succeed, and say precisely what its failure does and does not entitle you to conclude about an equation.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

$x = -8$ or $x = 3$ .
- $\{-8, 3\}$ is the same pair written as a set; reporting only one of the two values is not the same answer, since both satisfy the equation
Part B

$x = \dfrac{-5 + \sqrt{105}}{2}$ and $x = \dfrac{-5 - \sqrt{105}}{2}$ .
- $x = -\tfrac{5}{2} \pm \tfrac{\sqrt{105}}{2}$ is the same pair before the two terms are put over one denominator; a decimal such as $2.62$ is an approximation and not an exact value
Part C

The search draws on a finite list of integer pairs, so it succeeds only when the roots are rational. A failed search reports on the list that was searched, not on the equation: the second equation has two perfectly good real roots, and they are irrational, which is exactly what no integer pair could reach.
Worked solution

Part A

The Zero Product Property needs a zero on one side, so move the $24$ across before anything is set equal to zero.

$x^2 + 5x - 24 = 0 \quad \Longrightarrow \quad (x + 8)(x - 3) = 0$

Two numbers multiplying to $-24$ and adding to $5$ are $8$ and $-3$ . The product is zero, so a factor is zero, giving $x = -8$ or $x = 3$ . Both check in the original: $(-8)^2 + 5(-8) = 64 - 40 = 24$ , and $3^2 + 15 = 24$ .

Part B

Move the $20$ across and add the square of half the linear coefficient to both sides. Half of $5$ is $\tfrac{5}{2}$ , and $\left(\tfrac{5}{2}\right)^2 = \tfrac{25}{4}$ .

$\left(x + \frac{5}{2}\right)^2 = 20 + \frac{25}{4} = \frac{105}{4}$

The right side is positive, so take the square root of both sides with both signs, then subtract $\tfrac{5}{2}$ .

$x = -\frac{5}{2} \pm \frac{\sqrt{105}}{2} = \frac{-5 \pm \sqrt{105}}{2}$

Nothing in that route can fail, and the quadratic formula returns the same pair, as it must: the formula is this computation carried out once on the general equation.

Part C

Factoring over the integers is a search. For a monic $x^2 + bx + c$ you look for integers $p$ and $q$ with $p + q = b$ and $pq = c$ , and only finitely many pairs multiply to $c$ , which is why the hunt feels mechanical. It can succeed only when the roots are rational, because an integer pair hands back a factorization with integer coefficients, and that puts each root at a rational number.

$x^2 + 5x - 24 = (x + 8)(x - 3), \qquad x^2 + 5x - 20 = \left(x + \frac{5}{2}\right)^2 - \frac{105}{4}$

For the second, no integer pair multiplies to $-20$ and adds to $5$ , and none ever will, because the roots $\tfrac{-5 \pm \sqrt{105}}{2}$ are irrational. What follows from the failure is only that the list searched held no answer. The equation still has two real roots, and the route in part B, which guesses nothing, found them.

So a failed search is a report about the searcher's list, never about the equation, and in particular it is not the claim that there are no real solutions. That is a different claim with its own test, the sign of $b^2 - 4ac$ , which here is $25 + 80 = 105 > 0$ .

In one line

$x^2 + 5x = 24$ has the roots $x = -8$ and $x = 3$ , while $x^2 + 5x = 20$ has the roots $x = \tfrac{-5 \pm \sqrt{105}}{2}$ . The integer search finishes the first and can never finish the second, because the roots there are irrational, and that failure is a report about the list of candidate pairs rather than about the equation, which has two real roots throughout.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Gathers every term onto one side against zero before any factor is set equal to anything. . Worth 2 points.

Produces a correct factorization and solves both linear equations it yields. . Worth 2 points.

Reports both values as the solution set and substitutes at least one back into the original equation. . Worth 1 point.

Part B 5 points

Isolates a squared binomial with a bare constant on the other side, adding the completing constant to both sides. . Worth 2 points.

Keeps both signs when the square root is taken, and reaches a single exact expression for the pair. . Worth 2 points.

Reports both roots exactly rather than as rounded decimals. . Worth 1 point.

Part C 5 points

Locates the difference between the two equations in whether the roots are rational, and ties the search to the finite list of candidates rather than to the difficulty of the equation. . Worth 3 points. needs an explanation, not just an answer

Says what a failed search does not license, keeping no rational factorization apart from no real roots. . Worth 2 points. needs an explanation, not just an answer
3. Two curves that cross the axis in the same two places . 13 points. Question 3 of 10.

Two functions, $p(x) = 2x^2 - 9x - 5$ and $q(x) = 6x^2 - 27x - 15$ . They are not the same function. Neither root of either is needed anywhere below.
1. Part A.
  Give the sum and the product of the roots of $p$ , and the sum and the product of the roots of $q$ , without solving either equation.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  State the axis of symmetry of each parabola, and the $y$ -intercept of each.
  
  Carry your own answer forward Carry your own totals across from the previous part, and take each $y$ -intercept from the standard form printed in the stem. What earns credit is putting the axis at the midpoint of the roots and taking an intercept from the form that displays it.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Say whether the totals in part A had to come out the same for both quadratics, and explain why from the coefficients alone. Then say what a pair of roots therefore leaves undetermined about a quadratic.
  
  Carry your own answer forward Argue from your own totals, whatever they came out to be. What earns credit is the account you give of how the two quadratics' totals are related, and of what that relation leaves open, rather than the totals themselves.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

For $p$ : sum $\tfrac{9}{2}$ , product $-\tfrac{5}{2}$ . For $q$ : the same two totals, sum $\tfrac{9}{2}$ and product $-\tfrac{5}{2}$ .
- $4.5$ and $-2.5$ are the same two totals as decimals; a sum of $-\tfrac{9}{2}$ has kept the sign of $b$ instead of reversing it
Part B

Both have the axis $x = \tfrac{9}{4}$ . The $y$ -intercept of $p$ is $-5$ and the $y$ -intercept of $q$ is $-15$ .

Part C

They had to. Both totals are ratios, $-\tfrac{b}{a}$ and $\tfrac{c}{a}$ , and $q$ is $3p$ , so the factor of $3$ cancels out of each. A pair of roots therefore fixes a quadratic only up to a nonzero scalar: the leading coefficient stays free, and with it the $y$ -intercept and the vertex height.
Worked solution

Part A

Read the coefficients of each and divide both totals by the leading one, remembering that only the sum takes the extra minus sign.

$p: \ \ -\frac{-9}{2} = \frac{9}{2}, \ \ \frac{-5}{2}; \qquad q: \ \ -\frac{-27}{6} = \frac{9}{2}, \ \ \frac{-15}{6} = -\frac{5}{2}$

The two quadratics report the same pair of totals. Both have real roots to report on, since $\Delta_p = 81 + 40 = 121 > 0$ .

Part B

The two roots are a mirror pair about the axis, so the axis stands at their midpoint, which is half their sum.

$x = \frac{1}{2} \cdot \frac{9}{2} = \frac{9}{4} \quad \text{for both curves}$

Standard form hands over the $y$ -intercept for nothing, since $f(0) = c$ : it is $-5$ for $p$ and $-15$ for $q$ . The axis is shared because the roots are shared; the intercepts differ because one curve is three times the other.

Part C

Expanding the factored form and matching coefficients term by term is where the two totals come from:

$a(x - r_1)(x - r_2) = ax^2 - a(r_1 + r_2)x + a r_1 r_2 \quad \Longrightarrow \quad r_1 + r_2 = -\frac{b}{a}, \ \ r_1 r_2 = \frac{c}{a}$

Both are ratios with $a$ downstairs. Multiplying a quadratic by $3$ multiplies $a$ , $b$ and $c$ alike, so each ratio is unchanged and $q$ has to report exactly what $p$ reports.

Read backwards, that is a limit on what a pair of roots can tell you. Every member of the family $a(x - r_1)(x - r_2)$ has those same two roots, so naming the roots pins down the factors and nothing else. The leading coefficient is left free, and every vertical measurement rides on it: here the two curves share their crossings and their axis while their $y$ -intercepts differ by a factor of $3$ . To pin down a single parabola you need one further piece of information, such as one more point on the curve.

In one line

$p$ and $q$ report the same totals, sum $\tfrac{9}{2}$ and product $-\tfrac{5}{2}$ , and therefore share the axis $x = \tfrac{9}{4}$ , while their $y$ -intercepts differ, $-5$ against $-15$ . Both totals are ratios with the leading coefficient downstairs, so tripling every coefficient leaves them untouched: a pair of roots fixes a quadratic only up to a nonzero scalar, which is exactly what the two intercepts show.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Divides both totals by the leading coefficient and keeps the extra minus sign on the sum alone. . Worth 2 points.

Reports both pairs of totals and notes whether the two quadratics agree. . Worth 2 points.

Part B 4 points

Places the axis at half the sum of the roots rather than at the sum itself. . Worth 2 points.

Reads each $y$ -intercept off the constant term, and says which of the two features the curves share. . Worth 2 points.

Part C 5 points

Reaches a verdict on whether the totals must match, and supports it from both totals being ratios out of which the leading coefficient cancels, rather than from the two quadratics simply sharing their roots. . Worth 3 points. needs an explanation, not just an answer

States what a pair of roots leaves free, and names a feature that a nonzero scalar changes. . Worth 2 points. needs an explanation, not just an answer
4. A reservoir, and the months it holds its level . 13 points. Question 4 of 10.

The depth of water in a reservoir is modelled across one year by

$d(t) = t^2 - 9t + 26,$

where $d$ is the depth in metres and $t$ is the number of months after the first of January. The model is used only on its own window, $0 \le t \le 12$ .
1. Part A.
  Write the requirement that the depth is at least $8$ metres as an inequality with zero on one side, and find the two times at which the depth is exactly $8$ metres.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 4 points
2. Part B.
  Give the set of times in the model's window at which the depth is at least $8$ metres.
  
  Carry your own answer forward Work from the inequality and the boundary times you produced in part A, whatever they were. The credit here is for splitting the window at your own boundary times, for keeping the pieces the symbol asks for, and for the endpoints, not for one particular set.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Say which times in the window the model puts strictly below $8$ metres. Determine which of your two sets contains the boundary times, and explain why. Then say what would change in both answers if the requirement had been more than $8$ metres instead of at least $8$ metres.
  
  Carry your own answer forward Argue from your own part B set and your own boundary times. The credit here is for the account of which side of the answer an endpoint falls on and for the effect of changing the symbol, not for particular months.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
The answer

Part A

$t^2 - 9t + 18 \ge 0$ , and the depth is exactly $8$ metres at $t = 3$ and at $t = 6$ .
Part B

$[0, 3] \cup [6, 12]$ : the first three months of the year and the last six.
- $0 \le t \le 3$ or $6 \le t \le 12$ is the same set in words; a single interval such as $[3, 6]$ or $[0, 12]$ is a different set entirely
Part C

Strictly below on $(3, 6)$ . At the boundary times the depth is exactly $8$ metres, which at least accepts and below does not, so they sit in the part B set. A strict requirement would drop them, leaving $[0, 3) \cup (6, 12]$ , while the times strictly below $8$ metres would still be $(3, 6)$ .
Worked solution

Part A

Set the model against the threshold and gather every term onto one side.

$t^2 - 9t + 26 \ge 8 \quad \Longrightarrow \quad t^2 - 9t + 18 \ge 0 \quad \Longrightarrow \quad (t - 3)(t - 6) \ge 0$

The two times at which the depth is exactly $8$ metres are the roots of that quadratic, $t = 3$ and $t = 6$ , which is the first of April and the first of July.

Part B

The two boundary times cut the line into three pieces, and $t^2 - 9t + 18$ opens upward, so it is negative between its roots and positive outside them. The symbol is non-strict, so the roots themselves belong to the answer.

$t \le 3 \quad \text{or} \quad t \ge 6$

Only times inside the model's own window count, so intersect with $0 \le t \le 12$ :

$[0, 3] \cup [6, 12]$

The union is not decoration. The answer is two separate stretches of the year, and writing it as one interval would report a different set.

Part C

The complement of the part B set inside the window is the stretch between the boundary times, and it is open at both ends.

$\text{at least } 8: \ [0, 3] \cup [6, 12], \qquad \text{strictly below } 8: \ (3, 6)$

At $t = 3$ and $t = 6$ the depth is exactly $8$ metres. The phrase at least $8$ accepts equality, so those two instants belong to the first set; strictly below does not, so they are excluded from the second. Nothing about the interior of either set is in question, and only the endpoints move.

That is also what changes under a strict requirement. Asking for more than $8$ metres would remove the two instants from the first set, leaving $[0, 3) \cup (6, 12]$ . The times strictly below $8$ metres are untouched by that change and are still $(3, 6)$ ; what becomes the closed $[3, 6]$ is the set of times that FAIL the new requirement, which now takes in the two instants where the depth is exactly $8$ metres. The stretches are the same either way; the symbol decides only who owns the boundary.

In one line

The requirement is $t^2 - 9t + 18 \ge 0$ , with boundary times $t = 3$ and $t = 6$ , so inside the window the depth is at least $8$ metres on $[0, 3] \cup [6, 12]$ and strictly below $8$ metres on $(3, 6)$ . The boundary times belong to the first set because the depth there is exactly $8$ metres, which is what at least accepts; asking for more than $8$ metres would drop them from it, leaving $[0, 3) \cup (6, 12]$ , while the times strictly below $8$ metres would still be $(3, 6)$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Compares the model to the threshold and gathers every term onto one side against zero before anything else is done. . Worth 2 points.

Finds both boundary times correctly from the resulting quadratic. . Worth 2 points.

Part B 5 points

Keeps the pieces on which the quadratic is at or above zero, which for an upward parabola are the two outside the roots. . Worth 2 points.

Cuts the answer down to the model's own window and reports it as a union of two stretches rather than one interval. . Worth 2 points.

Says what the set means in months of the year. . Worth 1 point.

Part C 4 points

Names the complement inside the model's window and gives it the endpoints the strict reading demands. . Worth 2 points.

Explains the endpoint decision from what the symbol accepts at a root, and says what a strict requirement would move and what it would leave alone. . Worth 2 points. needs an explanation, not just an answer
5. A path around a bed, and one number computed before the answer . 15 points. Question 5 of 10.

A rectangular vegetable bed measures $8$ metres by $5$ metres. A path of uniform width is laid all the way around it, on all four sides, and the path alone covers $68$ square metres of ground.
1. Part A.
  Name the unknown and turn the sentence about the path's area into an equation in standard form.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 4 points
2. Part B.
  Compute the discriminant of your equation and say what it predicts, then solve, reporting both roots before deciding anything about the path.
  
  Carry your own answer forward Use whichever equation you produced in part A and finish honestly from it. The credit here is for computing the discriminant before solving, for a correctly formed substitution, and for testing each root against what the letter stands for.
  
  Solve and show your work Write each step out, and end with the value and its units. 6 points
3. Part C.
  The discriminant was computed in part B before either root was known. State what that one number settled and what it left completely open, and decide whether a solver who means to find the roots anyway has any reason to compute it first.
  
  Carry your own answer forward Argue from whichever discriminant and roots you produced in part B, even if they were not the expected ones. The credit here is for the account of what the number can and cannot settle, not for a particular verdict.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

With $w$ the width of the path in metres, $(8 + 2w)(5 + 2w) - 40 = 68$ , which in standard form is $2w^2 + 13w - 34 = 0$ .
Part B

The discriminant is $441$ , so there are two distinct real roots: $w = 2$ and $w = -\tfrac{17}{2}$ . A width cannot be negative, so the path is $2$ metres wide.
- $w = -8.5$ is the same rejected root as a decimal; an answer of $2$ metres with no mention of the second root has skipped the step where a root is tested against what the letter stands for
Part C

It settled how many real roots there are, being the quantity under the radical: $z^2 = \Delta$ has two real solutions, one, or none according to its sign. It left the values open, which need $-b$ and $2a$ too, and said nothing about which root a width can be. It is still worth the one subtraction first.
Worked solution

Part A

Let $w$ be the width of the path in metres. The path adds $w$ to each of the four sides, so it widens each dimension twice over: the outer rectangle is $(8 + 2w)$ by $(5 + 2w)$ . The path itself is what is left once the bed is removed.

$(8 + 2w)(5 + 2w) - 40 = 4w^2 + 26w = 68$

Gathering onto one side and dividing every term by $2$ gives the standard form $2w^2 + 13w - 34 = 0$ .

Part B

With $a = 2$ , $b = 13$ and $c = -34$ , compute the discriminant first: it costs one line and it says what kind of answer to expect.

$\Delta = 13^2 - 4(2)(-34) = 169 + 272 = 441$

It is positive, so there are two distinct real roots, and since $441 = 21^2$ the radical comes out whole. Substituting into the formula,

$w = \frac{-13 \pm 21}{4}, \qquad w = 2 \ \text{ or } \ w = -\frac{17}{2}.$

Both satisfy the equation. Only one of them is a width, since a length cannot be negative, so the path is $2$ metres wide. Checking it: the outer rectangle is $12$ by $9$ , the bed is $40$ square metres, and $108 - 40 = 68$ .

Part C

The discriminant is the quantity under the radical of $w = \tfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , so the three cases are the three things the equation $z^2 = \Delta$ can do: two real solutions when $\Delta > 0$ , exactly one when $\Delta = 0$ , and none at all when $\Delta < 0$ .

$\Delta = 441 > 0 \quad \Longrightarrow \quad \text{two distinct real roots}$

What it leaves open is everything else. The values of the roots need $-b$ and $2a$ as well, which the discriminant never sees, and the question of which root is a width is not a question about the equation at all: both roots satisfy it exactly, and $-\tfrac{17}{2}$ is rejected only because the letter was named as a length.

So yes, it is worth computing first even when you intend to solve. One subtraction tells you in advance whether the equation has real roots at all, which is the difference between a computation that is going somewhere and one that will end at the square root of a negative number.

In one line

With $w$ the path's width in metres the model is $2w^2 + 13w - 34 = 0$ , whose discriminant is $441$ , so there are two distinct real roots, $w = 2$ and $w = -\tfrac{17}{2}$ . The path is $2$ metres wide, the other root being a genuine solution of the equation that a width cannot be. The discriminant settled that count before any solving, and did not determine the individual root values by itself; $a$ and $b$ are also needed.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Widens both dimensions by twice the unknown, one width on each side, rather than by the width once. . Worth 2 points.

Subtracts the bed's own area so the equation is about the path alone, and reaches standard form. . Worth 2 points.

Part B 6 points

Computes the discriminant before solving and states what it predicts about the number of roots. . Worth 2 points.

Substitutes into a correctly written formula, the whole numerator over $2a$ , and simplifies. . Worth 2 points.

Reports both roots, rejects the inadmissible one with the reason named, and gives the width with its unit. . Worth 2 points.

Part C 5 points

Locates the discriminant as the quantity the formula takes a square root of, and ties the three cases to what a real square root can do rather than quoting them as a rule. . Worth 3 points. needs an explanation, not just an answer

Says what the discriminant leaves undetermined, separating the values of the roots from the count and from the modelling decision, and reaches a verdict on computing it first. . Worth 2 points. needs an explanation, not just an answer
6. A rule about empty solution sets . 13 points. Question 6 of 10.

A student writes down a rule: a quadratic inequality has an empty solution set exactly when the quadratic has no real root. The parts below put that rule under pressure from both sides.
1. Part A.
  Take $u(x) = 2x^2 - 3x + 8$ . Which real numbers satisfy $u(x) > 0$ , and which satisfy $u(x) < 0$ ?
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
2. Part B.
  Produce a quadratic and an inequality symbol for which the solution set is empty even though the quadratic does have a real root, and show that your example really does both of those things.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 4 points
3. Part C.
  Repair the rule. State exactly what an empty solution set turns on, and support your statement by giving, for a quadratic with no real root, one symbol that makes the set empty and one that makes it every real number.
  
  Carry your own answer forward Support the repair with whichever solution sets and example you produced in parts A and B, even if they were not the expected ones. The credit here is for the repaired statement and for a case in each direction, not for particular sets.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

$u(x) > 0$ has solution set $\mathbb{R}$ , and $u(x) < 0$ has solution set $\varnothing$ .

Part B

$(4x - 3)^2 < 0$ . The quadratic has the real root $x = \tfrac{3}{4}$ , where its value is zero, and a real square is never negative, so no real number satisfies the strict inequality.

Part C

Emptiness turns on three things together: the sign of the leading coefficient, the sign of the discriminant, and the symbol. For $u$ above, with no real root and $a > 0$ , the set is empty for $u(x) < 0$ and all of $\mathbb{R}$ for $u(x) > 0$ ; and part B shows a real root does not prevent emptiness either.

Worked solution

Part A

First find out whether there is any root at which the sign could turn.

$\Delta = (-3)^2 - 4(2)(8) = 9 - 64 = -55 < 0$

There is none, so $u$ keeps one sign across the whole line, and which sign follows from the leading coefficient: $a = 2 > 0$ and the vertex sits at height $-\tfrac{\Delta}{4a} = \tfrac{55}{8} > 0$ , so the lowest point of the curve is already above the axis and every output is positive.

Hence $u(x) > 0$ is satisfied by every real number, and $u(x) < 0$ by none.

Part B

Take $g(x) = (4x - 3)^2 = 16x^2 - 24x + 9$ , whose discriminant is $576 - 576 = 0$ . So $g$ has a real root, the repeated root $x = \tfrac{3}{4}$ , and the parabola is tangent to the axis there.

$g(x) < 0 \quad \Longrightarrow \quad (4x - 3)^2 < 0$

A real square is never negative, so nothing satisfies this and the solution set is $\varnothing$ . The rule fails in this direction: emptiness arrived with a real root present, because the symbol asked for something strictly negative while the least value the quadratic takes is zero.

Part C

The rule fails in both directions, and the repair names three ingredients where the rule named one.

$\text{empty} \ \Longleftrightarrow \ \text{the inequality asks for a sign the quadratic never takes}$

Which sign that is depends on the sign of $a$ , the sign of $\Delta$ and the symbol, all three. Part A supplies one direction: $u$ has no real root, yet $u(x) > 0$ holds for every real number and only $u(x) < 0$ is empty, so having no real root cannot by itself deliver emptiness. Part B supplies the other: $(4x - 3)^2 < 0$ is empty although the quadratic has a real root.

The sign of the leading coefficient is the third ingredient rather than a detail. If $u$ were replaced by $-u$ , which still has no real root, the two answers in part A would swap: $-u(x) < 0$ would be all of $\mathbb{R}$ and $-u(x) > 0$ would be empty.

In one line

For $u(x) = 2x^2 - 3x + 8$ the discriminant is $-55$ , so $u(x) > 0$ has solution set $\mathbb{R}$ while $u(x) < 0$ has solution set $\varnothing$ ; and $(4x - 3)^2 < 0$ is empty although that quadratic has the real root $\tfrac{3}{4}$ . The rule is therefore wrong in both directions: an empty solution set turns on the sign of the leading coefficient, the sign of the discriminant and the inequality symbol together, never on the root count alone.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Decides first whether any root exists, since a root is the only place the sign can turn. . Worth 2 points.

Gives both answers as sets, taking the constant sign from the leading coefficient or the vertex height rather than from sample points alone. . Worth 2 points.

Part B 4 points

Exhibits a quadratic with a real root together with a symbol that no real number can satisfy. . Worth 2 points.

Shows both halves of the example: that the root is real, and that nothing satisfies the inequality. . Worth 2 points. needs an explanation, not just an answer

Part C 5 points

Repairs the rule by naming every ingredient an empty solution set turns on, rather than any one of them alone. . Worth 3 points. needs an explanation, not just an answer

Supports the repair with a case in each direction, including a quadratic with no real root whose inequality is satisfied by every real number. . Worth 2 points. needs an explanation, not just an answer
7. A vertex placed to order . 14 points. Question 7 of 10.

Every member of the family $y = x^2 + 8x + c$ has the same shape and the same axis of symmetry, whatever the real number $c$ may be. Changing $c$ only slides the curve up or down.
1. Part A.
  Write the height of the vertex as an expression in $c$ , and find the value of $c$ that puts the vertex exactly $5$ units below the $x$ -axis.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Give every value of $c$ for which the curve has no real zero, and every value for which it has exactly one.
  
  Carry your own answer forward You may argue from the vertex height you produced in part A or from the discriminant directly; either route is fine, and if your part A expression was not the expected one, use it honestly. The credit here is for turning each root count into a condition and for the direction of the inequality.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  One number answered both parts above. Explain why the height of the vertex and the number of real zeros cannot be independent of each other for a curve of this shape, and say what changes in that account if the leading coefficient is negative.
  
  Carry your own answer forward Argue from the vertex height and the conditions you produced above, whatever they were. The credit here is for the account of why one number governs both questions, not for the particular values of $c$ .
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer
Part A

The vertex height is $c - 16$ , and it equals $-5$ when $c = 11$ .
- $-16 + c$ is the same expression; a height of $16 - c$ has the subtraction the wrong way round and would put the vertex above the axis for small $c$
Part B

No real zero when $c > 16$ ; exactly one real zero when $c = 16$ .
- $c \in (16, \infty)$ is the same set; an answer of $c \ge 16$ for no real zero has swallowed the boundary case, where there is one zero rather than none
Part C

The vertex is the lowest point of an upward parabola, so the curve meets the axis exactly when that point is at or below it: one number, the vertex height $-\tfrac{\Delta}{4a}$ , decides both. A negative leading coefficient makes the vertex the highest point, so the height takes the sign of $\Delta$ instead of the opposite one.
Worked solution

Part A

The height of the vertex is $-\tfrac{\Delta}{4a}$ , and $c$ enters the discriminant once.

$-\frac{\Delta}{4a} = -\frac{64 - 4c}{4} = c - 16$

The same expression comes from evaluating on the axis $x = -\tfrac{b}{2a} = -4$ , since $16 - 32 + c = c - 16$ . Below the axis means a negative height, so set $c - 16 = -5$ , which gives $c = 11$ .

Part B

Both questions are questions about the sign of one number, and $c$ enters it once.

$\Delta = 8^2 - 4(1)c = 64 - 4c$

No real zero is $\Delta < 0$ , that is $64 < 4c$ , so $c > 16$ . Exactly one real zero is $\Delta = 0$ , that is $c = 16$ , where the curve is tangent to the axis at $x = -4$ . The vertex height says the same thing in the picture: $c - 16$ is positive exactly when $c > 16$ , and an upward parabola whose lowest point is above the axis never reaches it.

Part C

For $a > 0$ the vertex is the lowest point of the graph, so the whole curve lies at or above that height. The curve can meet the $x$ -axis only if its lowest point is at or below the axis, and it must meet it if that point is strictly below, because the arms climb without bound on both sides. So the two questions are one question.

$\text{vertex height} = -\frac{\Delta}{4a} = c - 16$

With $a$ fixed and positive, the height and $\Delta$ carry opposite signs, so a vertex below the axis and a positive discriminant are the same statement, and a vertex on the axis and a zero discriminant are the same statement.

With $a < 0$ the vertex is the highest point instead, and $-\tfrac{\Delta}{4a}$ now carries the same sign as $\Delta$ , so it is a vertex above the axis that means two crossings. The three-way count itself does not move: it was always a statement about the sign of $\Delta$ alone, and only the picture that goes with each case is turned over.

In one line

The vertex of $y = x^2 + 8x + c$ sits at height $c - 16$ , so it is $5$ units below the axis when $c = 11$ ; the curve has no real zero when $c > 16$ and exactly one when $c = 16$ . One number governs both because for an upward parabola the vertex is the lowest point of the graph, and that height is $-\tfrac{\Delta}{4a}$ . A negative leading coefficient makes the vertex the highest point instead, so the height takes the sign of $\Delta$ rather than the opposite, while the rule that reads the number of real zeros off the sign of $\Delta$ is untouched.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Reaches the vertex height as an expression in $c$ , from $-\tfrac{\Delta}{4a}$ or by evaluating on the axis of symmetry. . Worth 2 points.

Reads below the axis as a negative height and solves for $c$ . . Worth 2 points.

Part B 5 points

Turns each root count into the matching condition on the discriminant, or on the vertex height, rather than testing sample values. . Worth 2 points.

Gets the direction of the inequality right, accounting for the sign with which $c$ enters. . Worth 2 points.

Reports the boundary value separately from the region and says what the curve does there. . Worth 1 point.

Part C 5 points

Argues from the vertex being the extreme point of the curve to the number of crossings, rather than quoting the three cases as a rule. . Worth 3 points. needs an explanation, not just an answer

Says what a negative leading coefficient changes and what it leaves exactly as it was. . Worth 2 points.
8. One route run twice, once with numbers and once without . 14 points. Question 8 of 10.

The equation $5x^2 + 6x - 3 = 0$ has a leading coefficient that does not divide its middle coefficient evenly, which is where the route below is most often botched.
1. Part A.
  Rewrite the left side as a multiple of a squared binomial plus a constant.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Solve the equation from your part A form, then check the pair you get against $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ and report whether the two agree.
  
  Carry your own answer forward Solve from your own completed form above, then run the check honestly and say what it reports. What earns credit is isolating the square, keeping both signs, and carrying the comparison out, rather than arriving at one particular pair.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Carry the same route out on the general equation $ax^2 + bx + c = 0$ , with $a \ne 0$ , far enough to show where the $b^2 - 4ac$ under the formula's radical comes from, and say what the result of doing so makes the formula: a rule to be remembered, or something else.
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 5 points
The answer
Part A

$5\left(x + \tfrac{3}{5}\right)^2 - \tfrac{24}{5}$ .
- $5\left(x + \tfrac{3}{5}\right)^2 - 4.8$ is the same constant as a decimal; $5\left(x + \tfrac{3}{5}\right)^2 - \tfrac{84}{25}$ is what comes of leaving the $-\tfrac{9}{25}$ inside the bracket unmultiplied
Part B

$x = \dfrac{-3 + 2\sqrt{6}}{5}$ and $x = \dfrac{-3 - 2\sqrt{6}}{5}$ , and the two routes agree.
- $x = -\tfrac{3}{5} \pm \tfrac{2\sqrt{6}}{5}$ is the same pair before the terms are put over one denominator; $\tfrac{-6 \pm 4\sqrt{6}}{10}$ is the same pair not yet in lowest terms
Part C

Dividing by $a$ and adding $\tfrac{b^2}{4a^2}$ to both sides gives $\left(x + \tfrac{b}{2a}\right)^2 = \tfrac{b^2 - 4ac}{4a^2}$ : the discriminant is what the right side becomes over a common denominator. The formula is that computation done once in general, so it is a theorem rather than a rule to remember.
Worked solution

Part A

Take the $5$ out of the $x^2$ and $x$ terms only; the constant carries no $x$ and stays outside the bracket.

$5x^2 + 6x - 3 = 5\left(x^2 + \frac{6}{5}x\right) - 3$

Inside, half of $\tfrac{6}{5}$ is $\tfrac{3}{5}$ and $\left(\tfrac{3}{5}\right)^2 = \tfrac{9}{25}$ , so add that and subtract it in the same line.

$5\left(\left(x + \frac{3}{5}\right)^2 - \frac{9}{25}\right) - 3 = 5\left(x + \frac{3}{5}\right)^2 - \frac{9}{5} - 3 = 5\left(x + \frac{3}{5}\right)^2 - \frac{24}{5}$

The $-\tfrac{9}{25}$ is inside the bracket, so the $5$ multiplies it too, which is the step the leading coefficient exists to trip.

Part B

Isolate the square, divide by the $5$ outside it, and take the root of both sides with both signs.

$5\left(x + \frac{3}{5}\right)^2 = \frac{24}{5} \ \Longrightarrow \ \left(x + \frac{3}{5}\right)^2 = \frac{24}{25} \ \Longrightarrow \ x = \frac{-3 \pm 2\sqrt{6}}{5}$

Now the check, with $a = 5$ , $b = 6$ , $c = -3$ . Simplifying the radical and cancelling the factor of $2$ common to every term of the fraction gives the same pair.

$x = \frac{-6 \pm \sqrt{36 + 60}}{10} = \frac{-6 \pm \sqrt{96}}{10} = \frac{-6 \pm 4\sqrt{6}}{10} = \frac{-3 \pm 2\sqrt{6}}{5}$

Part C

Run the same steps with letters in place of numbers. Dividing by $a$ is legal exactly because $a \ne 0$ , which is the one hypothesis the whole formula carries.

$x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \quad \Longrightarrow \quad x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a}$

The left side is now a perfect square by construction, and the right side goes over the common denominator $4a^2$ , writing $\tfrac{c}{a} = \tfrac{4ac}{4a^2}$ .

$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$

There is the discriminant, and it was not put there by hand: it is what the right side becomes. Multiplying both sides by $4a^2$ clears the denominator and leaves the square of a whole expression, which is the tidiest place to take a root.

$(2ax + b)^2 = \Delta$

A real square is never negative, so when $\Delta < 0$ there is no real solution at all. When $\Delta \ge 0$ the root comes with both signs, $2ax + b = \pm\sqrt{\Delta}$ , and dividing by $2a$ , which is nonzero, finishes the formula.

Because every step was carried out on arbitrary $a$ , $b$ and $c$ , the result is proved for every quadratic at once. That is what makes it a theorem: the work in parts A and B was never strictly necessary, because it had already been done in general.

In one line

$5x^2 + 6x - 3 = 5\left(x + \tfrac{3}{5}\right)^2 - \tfrac{24}{5}$ , so the roots are $x = \tfrac{-3 \pm 2\sqrt{6}}{5}$ , which is exactly what the formula returns. Running the same route on $ax^2 + bx + c = 0$ reaches $\left(x + \tfrac{b}{2a}\right)^2 = \tfrac{b^2 - 4ac}{4a^2}$ , so the discriminant is simply what the right side becomes over a common denominator, and the formula is that one computation carried out in general and kept.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Removes the leading coefficient from the $x^2$ and $x$ terms only, leaving the constant outside the bracket. . Worth 2 points.

Distributes the leading coefficient back across both the square and the constant that was subtracted inside it. . Worth 2 points.

Part B 5 points

Isolates the square by dividing by the coefficient outside it before any root is taken, and keeps both signs. . Worth 2 points.

Simplifies the radical and cancels a factor common to every term of the fraction, so the two routes can be compared as written. . Worth 2 points.

States whether the two routes agree, rather than leaving two unreconciled expressions side by side. . Worth 1 point.

Part C 5 points

Carries the same steps through in letters, dividing by the leading coefficient and adding the square of half the new linear coefficient to both sides. . Worth 2 points.

Shows the discriminant arriving as the right-hand side over a common denominator, and draws the conclusion that the result is proved once for every quadratic rather than remembered. . Worth 3 points. needs an explanation, not just an answer
9. Two quadratics one constant apart . 15 points. Question 9 of 10.

Two quadratics with integer coefficients, $m(x) = 10x^2 - 3x - 4$ and $n(x) = 10x^2 - 3x - 5$ , differ only in their constant terms.
1. Part A.
  Compute both discriminants, and say for each quadratic whether it factors into linear factors with rational coefficients. Give the factorization wherever there is one.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  A student looks at $n$ and says: its discriminant is positive, so $n$ must factor, and I simply cannot find the right pair of integers. Identify what has gone wrong in that reasoning, and say what the student's failed search does establish.
  
  Carry your own answer forward Judge the student's reasoning against whichever discriminant you computed for $n$ in part A. The credit here is for the distinction the student has missed, not for a particular value.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 5 points
3. Part C.
  State the test the parts above have been using, with every hypothesis it needs. Then decide this claim: since $\sqrt{2}\,x^2 + 3x + \sqrt{2}$ has discriminant $1$ , a perfect square, it factors into linear factors with rational coefficients.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

$\Delta_m = 169 = 13^2$ , so $m(x) = (5x - 4)(2x + 1)$ . $\Delta_n = 209$ , which is not a perfect square, so $n$ has no factorization with rational coefficients.
- $m(x) = 10\left(x - \tfrac{4}{5}\right)\left(x + \tfrac{1}{2}\right)$ is the same factorization with the leading coefficient pulled out; a claim that $n$ does not factor at all is a different and stronger claim than the one asked for
Part B

A positive discriminant promises two real roots and says nothing about which coefficients are allowed. Over the reals $n$ does factor; over the rationals it does not, and no further searching will change that, because $\sqrt{209}$ is irrational. The failed search establishes only that no pair on the list tried works.

Part C

The test holds for a quadratic whose coefficients are rational: such a quadratic factors over the rationals exactly when its discriminant is a perfect square. The claim is false, and the failure is the hypothesis, since $\sqrt{2}$ is not rational. The roots are $-\sqrt{2}$ and $-\tfrac{1}{\sqrt{2}}$ , neither of them rational.
Worked solution

Part A

Both have integer coefficients, so the test applies to both. Keep the sign of each negative constant inside the $-4ac$ term.

$\Delta_m = (-3)^2 - 4(10)(-4) = 9 + 160 = 169 = 13^2, \qquad \Delta_n = 9 + 200 = 209$

A perfect-square discriminant with rational coefficients means rational roots, and each rational root buys a factor with rational coefficients. For $m$ the roots are $\tfrac{3 \pm 13}{20}$ , that is $\tfrac{4}{5}$ and $-\tfrac{1}{2}$ , giving $m(x) = (5x - 4)(2x + 1)$ .

Now $209$ lies between $14^2 = 196$ and $15^2 = 225$ , so it is not a perfect square, $\sqrt{209}$ is irrational, and $n$ has no factorization with rational coefficients. It does factor over the reals, since a positive discriminant gives it two real roots.

Part B

Two different properties have been run together. A positive discriminant is exactly the condition for two distinct real roots, and each real root buys a real linear factor, so $n$ genuinely does factor over the reals:

$n(x) = 10\left(x - \frac{3 + \sqrt{209}}{20}\right)\left(x - \frac{3 - \sqrt{209}}{20}\right)$

What the student is looking for is a factorization with rational coefficients, and that demands strictly more. With rational coefficients it happens exactly when the discriminant is a perfect square, and $209$ is not one, so there is nothing to find and the search would have run out however long it went on.

What the failed search establishes on its own is only that no pair on the list tried works. A search reports on its own list; the discriminant reports on the quadratic.

Part C

The test, with its hypothesis attached: for a quadratic whose coefficients are rational, it factors into linear factors with rational coefficients exactly when its discriminant is a perfect square. Drop the hypothesis and the test says nothing at all.

The claim drops it. The discriminant really is $1$ :

$\Delta = 3^2 - 4\left(\sqrt{2}\right)\left(\sqrt{2}\right) = 9 - 8 = 1$

But $\sqrt{2}$ is not rational, so no conclusion about a rational factorization is available, and in fact there is none to be had:

$x = \frac{-3 \pm 1}{2\sqrt{2}}, \qquad x = -\frac{1}{\sqrt{2}} \ \text{ or } \ x = -\sqrt{2}$

Neither root is rational. The quadratic does factor over the reals, as $\sqrt{2}\left(x + \tfrac{1}{\sqrt{2}}\right)\left(x + \sqrt{2}\right)$ , which is the same distinction part B turned on. So the claim is false, and it is false precisely at the point where the hypothesis was dropped.

In one line

$\Delta_m = 169$ is a perfect square, so $m(x) = (5x - 4)(2x + 1)$ , while $\Delta_n = 209$ is not, so $n$ has two real roots and no rational factorization. A positive discriminant promises real roots and never promises a rational factorization, and the perfect-square test is a statement about quadratics with rational coefficients only: $\sqrt{2}\,x^2 + 3x + \sqrt{2}$ has discriminant $1$ and roots $-\sqrt{2}$ and $-\tfrac{1}{\sqrt{2}}$ , so it does not factor over the rationals.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Computes both discriminants correctly, keeping the sign of the negative constant term inside $-4ac$ . . Worth 2 points.

Decides each case by whether the discriminant is a perfect square, not by whether it is positive, and names the number system in the verdict. . Worth 2 points.

Gives a factorization whose expansion returns the quadratic it came from. . Worth 1 point.

Part B 5 points

Separates having real roots from factoring over the rationals, and says which of the two a positive discriminant actually delivers. . Worth 3 points. needs an explanation, not just an answer

States what the failed search does establish, keeping it a claim about the candidates rather than about the quadratic. . Worth 2 points.

Part C 5 points

States the test with its condition on the coefficients attached, rather than as a bare equivalence. . Worth 3 points. needs an explanation, not just an answer

Reaches a verdict on the claim and grounds it in whether the test's hypothesis is met, rather than in the discriminant computation. . Worth 2 points. needs an explanation, not just an answer
10. Two numbers named only by their sum and their product . 14 points. Question 10 of 10.

A sum and a product are two numbers, and a pair of roots is two numbers. The parts below pass between the two descriptions and ask what each one settles about the other.
1. Part A.
  Decide whether two real numbers exist with sum $9$ and product $14$ , and whether two real numbers exist with sum $9$ and product $25$ . Give the pair wherever one exists.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Suppose $-\tfrac{7}{2}$ is a root of $2x^2 + kx - 35$ . Name its second root and the value of $k$ , with no quadratic equation solved anywhere.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  The coefficients of a quadratic determine some quantities built from its two roots and not others. Explain what separates the two kinds, using $r_1^2 + r_2^2$ and $r_1 - r_2$ as your examples, and say exactly how much the coefficients do fix about the second of them.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

Sum $9$ and product $14$ : yes, the numbers are $2$ and $7$ . Sum $9$ and product $25$ : no real pair exists.
- $7$ and $2$ is the same pair in the other order; an answer that offers a pair for the second case has produced numbers that cannot both be real
Part B

The other root is $5$ , and $k = -3$ .
- the quadratic is then $2x^2 - 3x - 35$ ; a value of $k = 3$ has dropped the minus sign that the sum $-\tfrac{b}{a}$ carries
Part C

A quantity is fixed exactly when swapping the roots leaves it unchanged. $r_1^2 + r_2^2$ does, equalling $(r_1 + r_2)^2 - 2r_1r_2$ , so the coefficients fix it. $r_1 - r_2$ changes sign, so for two distinct roots only its square $\tfrac{\Delta}{a^2}$ is fixed; at $\Delta = 0$ the roots coincide and the difference is $0$ .
Worked solution

Part A

A real pair with sum $s$ and product $p$ exists exactly when $s^2 - 4p \ge 0$ , because such a pair, if it exists, is the pair of roots of $x^2 - sx + p$ , whose discriminant is $s^2 - 4p$ .

$9^2 - 4(14) = 81 - 56 = 25 \ge 0, \qquad 9^2 - 4(25) = 81 - 100 = -19 < 0$

The first passes, and the pair is the roots of $x^2 - 9x + 14 = (x - 2)(x - 7)$ , namely $2$ and $7$ , which do add to $9$ and multiply to $14$ . The second fails, so no two real numbers add to $9$ and multiply to $25$ : writing down $x^2 - 9x + 25$ does not conjure them into existence.

Part B

The product of the roots is $\tfrac{c}{a} = -\tfrac{35}{2}$ , and one root is known, so divide it out.

$r_2 = -\frac{35}{2} \div \left(-\frac{7}{2}\right) = 5$

Now use the other total. The roots sum to $-\tfrac{7}{2} + 5 = \tfrac{3}{2}$ , and the sum is $-\tfrac{k}{2}$ , so $k = -3$ . Both totals now have to agree, which is a free check: with $k = -3$ the quadratic is $2x^2 - 3x - 35 = (2x + 7)(x - 5)$ , whose roots are $-\tfrac{7}{2}$ and $5$ .

Part C

The sum and the product are everything the coefficients report, and both are unchanged when the two roots trade places, because $b$ and $c$ have no way of telling one root from the other. So a quantity built from both roots is determined by the coefficients exactly when it too survives that swap.

$r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1r_2 = \left(-\frac{b}{a}\right)^2 - \frac{2c}{a}$

That one survives, and the identity rewrites it in the two totals, so it is completely determined. The difference does not survive: swapping the roots turns $r_1 - r_2$ into $r_2 - r_1$ , a different number as soon as the two roots are distinct, and no expression in $b$ and $c$ can take two different values on one set of coefficients. What is determined is its square, which the swap leaves alone:

$(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1r_2 = \frac{b^2 - 4ac}{a^2} = \frac{\Delta}{a^2}$

So the coefficients fix how far apart the roots are and refuse to say which of them lies on the right, whenever there are two of them to tell apart. The one case where the difference itself is determined is $\Delta = 0$ : the roots coincide, $r_1 - r_2 = 0$ , and no sign is left to be ambiguous about. That is also why $\Delta \ge 0$ is exactly the real-root condition: a real squared gap cannot be negative.

In one line

Two real numbers with sum $9$ and product $14$ exist and are $2$ and $7$ , while none have sum $9$ and product $25$ , since $81 - 100 < 0$ . In $2x^2 + kx - 35$ the second root is $5$ and $k = -3$ . The coefficients determine exactly those quantities that survive swapping the two roots: $r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1r_2$ is one of them, while $r_1 - r_2$ is not: for two distinct roots only its square $\tfrac{\Delta}{a^2}$ is fixed, and the one case in which the difference itself is determined is $\Delta = 0$ , where the roots coincide and it is $0$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Tests each case against the condition $s^2 - 4p \ge 0$ rather than by hunting for numbers. . Worth 2 points.

Reaches a verdict in each case and produces the pair wherever the condition passes. . Worth 2 points.

Part B 5 points

Recovers the second root from one of the two totals rather than by solving the quadratic. . Worth 2 points.

Uses the other total to pin down $k$ , keeping the extra minus sign that the sum carries. . Worth 2 points.

Checks the pair against both totals, or against a factorization, before reporting. . Worth 1 point.

Part C 5 points

Identifies surviving a swap of the two roots as what separates the determined quantities from the undetermined ones. . Worth 3 points. needs an explanation, not just an answer

Says exactly what the coefficients do fix about the difference, naming its square and connecting that to the discriminant. . Worth 2 points. needs an explanation, not just an answer

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. Three costumes, and what each one hands over 12 points

Part A 4

Produces two linear factors whose expansion returns the original expression. 2
Locates the axis as the midpoint of the two zeros, or as $-\tfrac{b}{2a}$ , and reports it as a vertical line. 2

Part B 4

Takes each feature from a form that displays it, evaluating on the axis for the vertex rather than guessing its height. 2
Reports the range as an interval closed at the correct end, the end decided by the sign of the leading coefficient. 2

Part C 4

Names, for each of the three forms, a fact it hands over at sight, and grounds that in the structure of the form rather than in a remembered list. 3
Names, for each of the three forms, something it conceals. 1

2. Two equations that differ only in one number 15 points

Part A 5

Gathers every term onto one side against zero before any factor is set equal to anything. 2
Produces a correct factorization and solves both linear equations it yields. 2
Reports both values as the solution set and substitutes at least one back into the original equation. 1

Part B 5

Isolates a squared binomial with a bare constant on the other side, adding the completing constant to both sides. 2
Keeps both signs when the square root is taken, and reaches a single exact expression for the pair. 2
Reports both roots exactly rather than as rounded decimals. 1

Part C 5

Locates the difference between the two equations in whether the roots are rational, and ties the search to the finite list of candidates rather than to the difficulty of the equation. 3
Says what a failed search does not license, keeping no rational factorization apart from no real roots. 2

3. Two curves that cross the axis in the same two places 13 points

Part A 4

Divides both totals by the leading coefficient and keeps the extra minus sign on the sum alone. 2
Reports both pairs of totals and notes whether the two quadratics agree. 2

Part B 4

Places the axis at half the sum of the roots rather than at the sum itself. 2
Reads each $y$ -intercept off the constant term, and says which of the two features the curves share. 2

Part C 5

Reaches a verdict on whether the totals must match, and supports it from both totals being ratios out of which the leading coefficient cancels, rather than from the two quadratics simply sharing their roots. 3
States what a pair of roots leaves free, and names a feature that a nonzero scalar changes. 2

4. A reservoir, and the months it holds its level 13 points

Part A 4

Compares the model to the threshold and gathers every term onto one side against zero before anything else is done. 2
Finds both boundary times correctly from the resulting quadratic. 2

Part B 5

Keeps the pieces on which the quadratic is at or above zero, which for an upward parabola are the two outside the roots. 2
Cuts the answer down to the model's own window and reports it as a union of two stretches rather than one interval. 2
Says what the set means in months of the year. 1

Part C 4

Names the complement inside the model's window and gives it the endpoints the strict reading demands. 2
Explains the endpoint decision from what the symbol accepts at a root, and says what a strict requirement would move and what it would leave alone. 2

5. A path around a bed, and one number computed before the answer 15 points

Part A 4

Widens both dimensions by twice the unknown, one width on each side, rather than by the width once. 2
Subtracts the bed's own area so the equation is about the path alone, and reaches standard form. 2

Part B 6

Computes the discriminant before solving and states what it predicts about the number of roots. 2
Substitutes into a correctly written formula, the whole numerator over $2a$ , and simplifies. 2
Reports both roots, rejects the inadmissible one with the reason named, and gives the width with its unit. 2

Part C 5

Locates the discriminant as the quantity the formula takes a square root of, and ties the three cases to what a real square root can do rather than quoting them as a rule. 3
Says what the discriminant leaves undetermined, separating the values of the roots from the count and from the modelling decision, and reaches a verdict on computing it first. 2

6. A rule about empty solution sets 13 points

Part A 4

Decides first whether any root exists, since a root is the only place the sign can turn. 2
Gives both answers as sets, taking the constant sign from the leading coefficient or the vertex height rather than from sample points alone. 2

Part B 4

Exhibits a quadratic with a real root together with a symbol that no real number can satisfy. 2
Shows both halves of the example: that the root is real, and that nothing satisfies the inequality. 2

Part C 5

Repairs the rule by naming every ingredient an empty solution set turns on, rather than any one of them alone. 3
Supports the repair with a case in each direction, including a quadratic with no real root whose inequality is satisfied by every real number. 2

7. A vertex placed to order 14 points

Part A 4

Reaches the vertex height as an expression in $c$ , from $-\tfrac{\Delta}{4a}$ or by evaluating on the axis of symmetry. 2
Reads below the axis as a negative height and solves for $c$ . 2

Part B 5

Turns each root count into the matching condition on the discriminant, or on the vertex height, rather than testing sample values. 2
Gets the direction of the inequality right, accounting for the sign with which $c$ enters. 2
Reports the boundary value separately from the region and says what the curve does there. 1

Part C 5

Argues from the vertex being the extreme point of the curve to the number of crossings, rather than quoting the three cases as a rule. 3
Says what a negative leading coefficient changes and what it leaves exactly as it was. 2

8. One route run twice, once with numbers and once without 14 points

Part A 4

Removes the leading coefficient from the $x^2$ and $x$ terms only, leaving the constant outside the bracket. 2
Distributes the leading coefficient back across both the square and the constant that was subtracted inside it. 2

Part B 5

Isolates the square by dividing by the coefficient outside it before any root is taken, and keeps both signs. 2
Simplifies the radical and cancels a factor common to every term of the fraction, so the two routes can be compared as written. 2
States whether the two routes agree, rather than leaving two unreconciled expressions side by side. 1

Part C 5

Carries the same steps through in letters, dividing by the leading coefficient and adding the square of half the new linear coefficient to both sides. 2
Shows the discriminant arriving as the right-hand side over a common denominator, and draws the conclusion that the result is proved once for every quadratic rather than remembered. 3

9. Two quadratics one constant apart 15 points

Part A 5

Computes both discriminants correctly, keeping the sign of the negative constant term inside $-4ac$ . 2
Decides each case by whether the discriminant is a perfect square, not by whether it is positive, and names the number system in the verdict. 2
Gives a factorization whose expansion returns the quadratic it came from. 1

Part B 5

Separates having real roots from factoring over the rationals, and says which of the two a positive discriminant actually delivers. 3
States what the failed search does establish, keeping it a claim about the candidates rather than about the quadratic. 2

Part C 5

States the test with its condition on the coefficients attached, rather than as a bare equivalence. 3
Reaches a verdict on the claim and grounds it in whether the test's hypothesis is met, rather than in the discriminant computation. 2

10. Two numbers named only by their sum and their product 14 points

Part A 4

Tests each case against the condition $s^2 - 4p \ge 0$ rather than by hunting for numbers. 2
Reaches a verdict in each case and produces the pair wherever the condition passes. 2

Part B 5

Recovers the second root from one of the two totals rather than by solving the quadratic. 2
Uses the other total to pin down $k$ , keeping the extra minus sign that the sum carries. 2
Checks the pair against both totals, or against a factorization, before reporting. 1

Part C 5

Identifies surviving a swap of the two roots as what separates the determined quantities from the undetermined ones. 3
Says exactly what the coefficients do fix about the difference, naming its square and connecting that to the discriminant. 2

Multiple choice

Free response