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Complex Numbers and the Quadratic Formula: 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, 115 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. Two parts that keep their distance, and one operation that makes them meet . 9 points. Question 1 of 10.

Let $z = 5 - 3i$ and $w = -2 + 7i$ . Every answer below should end in standard form $a + bi$ , with the real part first and a single imaginary term second.
1. Part A.
  Compute $z + w$ and $z - w$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Compute $zw$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  A classmate looks at your sum from part A and says it can be simplified further, on the grounds that both of its terms are just numbers. Explain why the two parts of a complex number cannot be merged into one. Then explain why, despite that, the imaginary parts of $z$ and $w$ were able to change the REAL part of your answer in part B.
  
  Carry your own answer forward Argue from whichever results you reached in parts A and B, even if they were not the expected ones. The credit here is for the account of why the parts behave differently under the two operations, not for reproducing one particular pair of numbers.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer
Part A

$z + w = 3 + 4i$ and $z - w = 7 - 10i$ .
- $3 + 4i$ may be written $4i + 3$ , though standard form puts the real part first; what is not the same is $7$ or $7i$ for the sum, which merges two parts that stay separate
Part B

$zw = 11 + 41i$ .
- $11 + 41i$ only; an answer of $-10 + 41i$ has left the $i^2$ term unresolved, and $-31 + 41i$ has read $i^2$ as $+1$
Part C

The two parts cannot merge because $i$ is not a real number, so $3$ and $4i$ are unlike terms, exactly as $3$ and $4x$ are. They still reach the real part in a product, because multiplying two imaginary terms produces $i^2$ , and $i^2 = -1$ is real.
Worked solution

Part A

Combine real with real and imaginary with imaginary; for the difference the minus reaches both parts of $w$ .

$z + w = (5 + (-2)) + (-3 + 7)i = 3 + 4i, \qquad z - w = (5 - (-2)) + (-3 - 7)i = 7 - 10i.$

Part B

Multiply as two binomials, then replace $i^2$ with $-1$ , which turns $-21i^2$ into $+21$ .

$(5 - 3i)(-2 + 7i) = -10 + 35i + 6i - 21i^2 = 11 + 41i.$

Part C

Why the parts stay apart. The number $3$ counts units of $1$ and $4i$ counts units of $i$ , and $i$ is not real, so they are unlike terms, exactly as $3$ and $4x$ are:

$3 + 4i \ne 7, \qquad 3 + 4i \ne 7i.$

Why they interact under multiplication. Addition never multiplies two imaginary terms, but a product does, and $i \cdot i = i^2 = -1$ is real. In part B the term $(-3i)(7i) = 21$ landed in the real part: the imaginary parts did not become real, their product did.

In one line

$z + w = 3 + 4i$ , $z - w = 7 - 10i$ , and $zw = 11 + 41i$ . The two parts cannot be merged because $i$ is not real, so they are unlike terms; but a product multiplies two imaginary terms together, and $i^2 = -1$ is real, which is how the imaginary parts reach the real part of the answer.

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 3 points

Combines like parts and distributes the minus sign across both parts of the second number in the difference. . Worth 2 points.

Reports each result in standard form, one real part and one imaginary term. . Worth 1 point.

Part B 3 points

Expands the product completely and resolves the $i^2$ term into a real contribution with the correct sign. . Worth 2 points.

Collects the terms into standard form, one real part and one imaginary part. . Worth 1 point.

Part C 3 points

Grounds the reason the parts cannot combine in $i$ not being a real number, rather than in a rule that is simply not allowed. . Worth 2 points. needs an explanation, not just an answer

Locates the interaction in the product of two imaginary terms and names $i^2 = -1$ as what turns it real. . Worth 1 point.
2. A square, a constant, and a verdict you can read without solving . 11 points. Question 2 of 10.

This question is about the quadratic $x^2 + 10x + 41$ and the equation $x^2 + 10x + 41 = 0$ it produces. Because $x$ appears in both terms, nothing can be undone until it is gathered into one place.
1. Part A.
  Rewrite $x^2 + 10x + 41$ as a squared binomial plus a constant.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Solve $x^2 + 10x + 41 = 0$ , giving both roots in the form $a + bi$ .
  
  Carry your own answer forward You may continue from the form you produced in part A, or start again from the equation; either route is fine. If your part A form was not the expected one, use it anyway and solve honestly from it, because the credit here is for isolating the square, keeping both signs, and handling what is under the root.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Your form from part A shows the quadratic as a square with something added to it. Argue from that form alone, without solving and without reference to the roots you found, that no real number can satisfy $x^2 + 10x + 41 = 0$ .
  
  Carry your own answer forward Run the argument on whichever completed form you produced in part A, even if it was not the expected one, and say honestly what it does or does not rule out. The credit is for the reasoning about a real square, not for a particular constant.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer
Part A

$(x + 5)^2 + 16$ .
- $(x + 5)^2 + 16$ is the answer; $(x + 5)^2 + 41$ has kept the original constant without paying for the $25$ that was borrowed
Part B

$x = -5 + 4i$ and $x = -5 - 4i$ .
- $x = -5 \pm 4i$ is the same pair on one line; reporting only $x = -5 + 4i$ is not, because a square root carries two signs
Part C

For real $x$ the square $(x + 5)^2$ is never negative, so $(x + 5)^2 + 16$ is a positive number added to a nonnegative one and is therefore always positive. Something always positive is never zero, so no real $x$ solves the equation.
Worked solution

Part A

Half of $10$ is $5$ , and $5^2 = 25$ completes the square. Add and subtract it on the one side, then combine constants, since $-25 + 41 = 16$ .

$x^2 + 10x + 41 = \left(x^2 + 10x + 25\right) - 25 + 41 = (x + 5)^2 + 16.$

Part B

Now there is a second side, so move $41$ across, add $25$ to both sides, and take the root of both signs with $\sqrt{-16} = 4i$ .

$(x + 5)^2 = -16 \quad\Longrightarrow\quad x + 5 = \pm 4i \quad\Longrightarrow\quad x = -5 \pm 4i.$

Part C

The completed form is $(x + 5)^2 + 16$ . For any real $x$ the square $(x + 5)^2$ is never negative, so adding $16$ keeps the value at least $16$ :

$(x + 5)^2 + 16 \ge 16 > 0.$

A quantity that is always strictly positive is never zero, so the equation has no real solution. The argument never took a root and never mentioned $i$ .

In one line

$x^2 + 10x + 41 = (x + 5)^2 + 16$ , and the equation $x^2 + 10x + 41 = 0$ has the conjugate roots $x = -5 \pm 4i$ . The completed form settles the real case on its own: a real square is never negative, so $(x + 5)^2 + 16$ is always at least $16$ and never zero.

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 3 points

Uses the square of half the coefficient of $x$ , adding and subtracting it on the same side rather than on a side that does not exist. . Worth 2 points.

Reports a single squared binomial plus one combined constant, with the halved coefficient inside. . 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.

Takes the square root with both signs and converts the root of the negative into $i$ times a real square root. . Worth 2 points.

Reports both roots in the requested $a + bi$ form, real and imaginary parts separated. . Worth 1 point.

Part C 3 points

Argues from a real square never being negative to the whole expression never vanishing, not from the roots or the discriminant. . Worth 3 points. needs an explanation, not just an answer
3. Two high powers, and a product worth looking at twice . 11 points. Question 3 of 10.

Powers of $i$ do not grow; they go around. This question builds a complex number out of two of them and then multiplies it by its conjugate.
1. Part A.
  Evaluate $i^{75}$ and $i^{46}$ , and write $z = i^{75} + i^{46}$ in standard form.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Write down $\overline{z}$ for your $z$ from part A, then compute the product $z\overline{z}$ .
  
  Carry your own answer forward Use whichever $z$ you produced in part A, even if it was not the expected one. The credit here is for conjugating the right part of it and for multiplying the pair out correctly, not for landing on a particular number.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Your product in part B came out with no imaginary part at all. Show that this is guaranteed for every complex number and not a feature of this one, by working with a general $z = a + bi$ . Your argument should account for the sign that decides whether the result is a sum or a difference of two squares.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer
Part A

$i^{75} = -i$ , $i^{46} = -1$ , and $z = -1 - i$ .
- $z = -1 - i$ may be written $-1 - 1i$ ; what is not the same is $-1i - 1$ left unordered, or any answer that has multiplied an exponent by $i$
Part B

$\overline{z} = -1 + i$ , and $z\overline{z} = 2$ .

Part C

For $z = a + bi$ , the middle terms of $z\overline{z}$ are $-abi$ and $+abi$ , which cancel, so nothing imaginary can survive. What is left is $a^2 - (bi)^2$ , and since $i^2 = -1$ the subtracted $b^2i^2$ turns into an added $b^2$ , giving $a^2 + b^2$ , a sum.
Worked solution

Part A

Each power is fixed by the remainder of its exponent on division by $4$ . Since $75 = 4 \times 18 + 3$ and $46 = 4 \times 11 + 2$ :

$i^{75} = i^3 = -i, \qquad i^{46} = i^2 = -1, \qquad z = -i + (-1) = -1 - i.$

Part B

Conjugating flips the sign of the imaginary part only, so $\overline{z} = -1 + i$ . The two middle terms cancel:

$z\overline{z} = (-1 - i)(-1 + i) = 1 - i^2 = 2.$

Part C

Take $z = a + bi$ , so $\overline{z} = a - bi$ , and expand:

$z\overline{z} = (a + bi)(a - bi) = a^2 - abi + abi - b^2i^2 = a^2 + b^2.$

The middle terms $-abi$ and $+abi$ are opposites for any $a$ and $b$ , so nothing imaginary survives. And $-b^2i^2 = +b^2$ because $i^2 = -1$ , so the squares add.

In one line

$i^{75} = -i$ and $i^{46} = -1$ , so $z = -1 - i$ ; its conjugate is $\overline{z} = -1 + i$ and $z\overline{z} = 2$ . For any $z = a + bi$ the middle terms $-abi$ and $+abi$ cancel, so nothing imaginary survives, and the $i^2$ turns $a^2 - b^2i^2$ into $a^2 + b^2$ , a sum of two squares.

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 3 points

Reduces each exponent using its remainder on division by $4$ and reports both powers with their signs. . Worth 2 points.

Assembles the two results into standard form, real and imaginary values in their own parts. . Worth 1 point.

Part B 4 points

Forms the conjugate by changing the sign of the imaginary part only, leaving the real part untouched. . Worth 1 point.

Multiplies the pair out and resolves the $i^2$ correctly, leaving no imaginary part. . Worth 2 points.

Notes out loud that the product has come out with no imaginary part. . Worth 1 point.

Part C 4 points

Argues in general letters that the two middle terms are opposites that cancel, so no imaginary part survives for any complex number. . Worth 3 points. needs an explanation, not just an answer

Shows the $i^2$ turns the subtracted square into an added one, and states which form the result takes. . Worth 1 point.
4. One number, computed first, that names the answer before you have it . 11 points. Question 4 of 10.

Consider the equation $2x^2 + 9 = 6x$ . Its coefficients cannot be read off until everything is on one side with zero on the other.
1. Part A.
  Put the equation in standard form, state $a$ , $b$ , and $c$ with their signs, and compute the discriminant. Say what kind of roots it predicts.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Solve the equation, giving both roots in the form $a + bi$ .
  
  Carry your own answer forward Use the standard form and the discriminant you produced in part A, whatever they were, and finish honestly from them. The credit here is for a correctly formed substitution, for handling whatever your discriminant turns out to be under the root, and for dividing the whole numerator.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  In part A you named the kind of roots before you had found a single one of them. Explain what the discriminant is doing that lets one number answer a question about the roots without producing them, and state precisely what it leaves undetermined.
  
  Carry your own answer forward Explain this from the role the quantity plays in the formula, using whatever discriminant you computed in part A. The credit is for the account of why one number can settle the question, not for a particular verdict.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

Standard form $2x^2 - 6x + 9 = 0$ , with $a = 2$ , $b = -6$ , $c = 9$ . The discriminant is $-36$ , which is negative, so the equation has two complex conjugate roots.
Part B

$x = \dfrac{3}{2} + \dfrac{3}{2}i$ and $x = \dfrac{3}{2} - \dfrac{3}{2}i$ .
- $x = \frac{3 \pm 3i}{2}$ is the same pair before splitting the fraction into parts; the unreduced $\frac{6 \pm 6i}{4}$ is the same two numbers but is not in lowest terms
Part C

The discriminant is the quantity the formula is about to take a square root of, and the three cases are just the three things a square root can do: give a nonzero real, give zero, or turn imaginary. It fixes only the KIND of roots; the values need $-b$ and $2a$ , which it never sees.
Worked solution

Part A

Subtract $6x$ so the right side is zero, then read the coefficients and the discriminant.

$2x^2 - 6x + 9 = 0, \quad a = 2,\ b = -6,\ c = 9, \quad b^2 - 4ac = 36 - 72 = -36.$

The discriminant is negative, so the roots are a complex conjugate pair, settled before any solving.

Part B

Substitute, using $-b = 6$ and $\sqrt{-36} = 6i$ , then reduce the whole fraction by $2$ .

$x = \frac{6 \pm \sqrt{-36}}{4} = \frac{6 \pm 6i}{4} = \frac{3}{2} \pm \frac{3}{2}i.$

Part C

The discriminant is the quantity under the radical of $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , and the three cases are the three things a square root can do: a nonzero real gives two real roots, zero gives one repeated, a negative gives a conjugate pair.

$b^2 - 4ac = -36 < 0 \quad\Longrightarrow\quad \text{a complex conjugate pair}.$

It fixes only the kind of roots. Their values need $-b$ and $2a$ , which the discriminant never sees.

In one line

In standard form the equation is $2x^2 - 6x + 9 = 0$ , with $a = 2$ , $b = -6$ , $c = 9$ and discriminant $-36$ , so the roots are a complex conjugate pair: $x = \frac{3}{2} \pm \frac{3}{2}i$ . The discriminant settles the kind of roots because it is exactly what the formula takes a square root of, and it settles nothing about their values, which need $-b$ and $2a$ as well.

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 3 points

Reaches standard form before reading any coefficient, and reports all three coefficients with their signs. . Worth 2 points.

Evaluates the discriminant and names the kind of roots it predicts, not just the number. . Worth 1 point.

Part B 5 points

Substitutes into a correctly written formula, the whole numerator over $2a$ with the sign of $-b$ handled. . Worth 2 points.

Handles the root of the discriminant, then reduces by a factor common to every term of the fraction. . Worth 2 points.

Reports both roots in the requested $a + bi$ form, consistently with part A's verdict. . Worth 1 point.

Part C 3 points

Locates the discriminant as the quantity the formula square-roots and ties the three cases to what a square root can do, rather than quoting them as a rule. . Worth 2 points. needs an explanation, not just an answer

Says what the discriminant leaves undetermined: the values of the roots need information it does not carry. . Worth 1 point.
5. A sail, and a number the algebra offers that the world will not take . 11 points. Question 5 of 10.

A triangular sail is cut so that its height is $3$ feet less than twice its base, and the finished sail has an area of $27$ square feet. The area of a triangle is half the base times the height.
1. Part A.
  Name the unknown, write every quantity the situation mentions in terms of it, and turn the sentence about the 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. 3 points
2. Part B.
  Solve your equation, report both roots, and answer the question the situation is actually asking: what are the base and the height of the sail?
  
  Carry your own answer forward Solve whichever equation you produced in part A, and interpret its roots against the sail honestly, even if the equation was not the expected one. The credit here is for solving correctly, for testing each root against what the letter stands for, and for answering in the terms the question asked.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Explain what it means for a number to satisfy the equation but not the situation, and apply that distinction to each of the two roots you found, taking them one at a time. Then say whether a solver should carry "a word problem always discards one root" into the next problem they meet.
  
  Carry your own answer forward Apply the distinction to whichever roots you found, and to the reasons you gave for keeping or rejecting each of them. The credit is for the account of what the two tests are and for the verdict on the general rule, not for a particular number.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

With $b$ the base in feet, the height is $2b - 3$ feet and the area sentence is $\tfrac{1}{2}b(2b - 3) = 27$ , which in standard form is $2b^2 - 3b - 54 = 0$ .

Part B

The roots are $b = 6$ and $b = -\dfrac{9}{2}$ . Only $b = 6$ is a length, so the sail has a base of $6$ feet and a height of $9$ feet.

Part C

The rejected root satisfies the equation exactly; it fails only the meaning attached to the letter, since the model was built assuming $b$ is a length and lengths are not negative. And no: the reject step is a test each root must be put to, not a quota. Some situations admit both roots.

Worked solution

Part A

Let $b$ be the base in feet; the height is $2b - 3$ . Half the base times the height is $27$ , so clear the fraction and gather to standard form.

$\frac{1}{2}b(2b - 3) = 27 \quad\Longrightarrow\quad 2b^2 - 3b - 54 = 0.$

Part B

Read the coefficients $2$ , $-3$ , $-54$ ; the negative constant makes the $-4ac$ term add.

$b = \frac{3 \pm \sqrt{9 + 432}}{4} = \frac{3 \pm 21}{4} = 6 \text{ or } -\frac{9}{2}.$

A base cannot be negative, so keep $b = 6$ . The height is $2(6) - 3 = 9$ , so the sail is $6$ feet by $9$ feet.

Part C

Substitute $b = -\frac{9}{2}$ and the equation vanishes, so it is a genuine root:

$2\left(-\frac{9}{2}\right)^2 - 3\left(-\frac{9}{2}\right) - 54 = 54 - 54 = 0.$

What it fails is the meaning of the letter: naming $b$ a base in feet restricted the answers to positive numbers, and the algebra never knew that. So no general rule follows. Rejecting a root is a test applied to each root, not a quota, and some situations, like a thrown object passing a height twice, admit both.

In one line

With $b$ the base in feet, the model is $2b^2 - 3b - 54 = 0$ , whose roots are $b = 6$ and $b = -\frac{9}{2}$ . The sail has a base of $6$ feet and a height of $9$ feet, the second root being rejected because a base is a length and a length cannot be negative. That rejection is a test applied to each root, not a quota: a situation is free to admit both.

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 3 points

Names the unknown and expresses the second dimension in that same unknown, no second letter. . Worth 2 points.

Translates the area sentence into an equation and rearranges to standard form. . Worth 1 point.

Part B 4 points

Solves the quadratic correctly, handling the sign of the $-4ac$ term when the constant is negative, and reports both roots first. . Worth 2 points.

Tests each root against what the letter stands for and rejects the impossible one with the reason named. . Worth 1 point.

Answers the question asked, giving both dimensions in a sentence with units. . Worth 1 point.

Part C 4 points

Distinguishes satisfying the equation from answering the question, locating the rejection in the meaning of the letter rather than the algebra. . Worth 2 points. needs an explanation, not just an answer

Reaches a verdict on the general rule and supports it with a specific situation, accounting for each of its roots. . Worth 2 points. needs an explanation, not just an answer
6. Two quotients, and the two jobs one move has to do . 12 points. Question 6 of 10.

A quotient of complex numbers is not finished while an $i$ is still sitting downstairs. Neither quotient below is in standard form yet, and the second has a denominator with no real part at all.
1. Part A.
  Write $\dfrac{7 - i}{3 + i}$ in the form $a + bi$ , and verify your answer by multiplying it back by the denominator.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Write $\dfrac{4 - 5i}{2i}$ in the form $a + bi$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  Multiplying the top and the bottom of a quotient by the conjugate of the denominator is doing two separate jobs at once, and the answer would be wrong if either one failed. Name both jobs and say what makes each one work. Then explain what would go wrong if you multiplied top and bottom by the conjugate of the NUMERATOR instead.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

$2 - i$ , and $(2 - i)(3 + i) = 7 - i$ , the original numerator.
- $2 - i$ may be written $2 - 1i$ ; an answer left as $\frac{20 - 10i}{10}$ is the same number but is not yet in standard form
Part B

$-\dfrac{5}{2} - 2i$ .
- $-2.5 - 2i$ is the same number; $\frac{-10 - 8i}{4}$ is the same number unreduced and not yet in standard form
Part C

One job is to leave the value alone, since the multiplier $\frac{\overline{w}}{\overline{w}}$ is $1$ ; the other is to clear the $i$ downstairs, since $w\overline{w} = c^2 + d^2$ is real. The numerator's conjugate does the first and not the second: the value survives, but the denominator stays complex.
Worked solution

Part A

Multiply top and bottom by the denominator's conjugate $3 - i$ , so the denominator becomes $3^2 + 1^2 = 10$ .

$\frac{7 - i}{3 + i} = \frac{(7 - i)(3 - i)}{10} = \frac{20 - 10i}{10} = 2 - i.$

Check by multiplying back: $(2 - i)(3 + i) = 7 - i$ , the original numerator.

Part B

Treat $2i$ as $0 + 2i$ ; its conjugate is $-2i$ , and $(2i)(-2i) = 4$ .

$\frac{4 - 5i}{2i} = \frac{(4 - 5i)(-2i)}{4} = \frac{-10 - 8i}{4} = -\frac{5}{2} - 2i.$

The numerator's $10i^2$ became $-10$ , which put a real part back on top.

Part C

Write the quotient as $\dfrac{z}{w}$ with $w = c + di$ ; the move multiplies by $\dfrac{\overline{w}}{\overline{w}}$ , doing two independent jobs. First it preserves the value, because $\dfrac{\overline{w}}{\overline{w}} = 1$ , and any nonzero multiplier equal to $1$ would do that. Second it clears the $i$ , because

$w\overline{w} = (c + di)(c - di) = c^2 + d^2$

is real, which the conjugate specifically guarantees. Multiplying instead by the numerator's conjugate $\overline{z}$ still equals $1$ , so the value survives, but the denominator $w\overline{z}$ is not a conjugate pair and stays complex: a legal line that makes no progress.

In one line

$\dfrac{7 - i}{3 + i} = 2 - i$ and $\dfrac{4 - 5i}{2i} = -\dfrac{5}{2} - 2i$ . Multiplying by the denominator's conjugate over itself does two jobs: it preserves the value, because the multiplier equals $1$ , and it clears the $i$ downstairs, because $w\overline{w} = c^2 + d^2$ is real. The numerator's conjugate does the first and not the second, leaving a legal line that makes no progress.

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

Multiplies numerator and denominator by the conjugate of the denominator, produces a real denominator, and resolves the $i^2$ . . Worth 2 points.

Divides the whole numerator to standard form and checks by multiplying back against the original numerator. . Worth 2 points.

Part B 3 points

Treats the pure imaginary denominator as a complex number with real part zero and clears the $i$ from downstairs. . Worth 2 points.

Resolves the $i^2$ , divides both parts, reduces, and reports standard form. . Worth 1 point.

Part C 5 points

Separates the two jobs with the right reason for each: the multiplier is a fraction equal to $1$ so the value is preserved, and the conjugate product is real so the denominator is cleared. . Worth 3 points. needs an explanation, not just an answer

Works out what the numerator's conjugate does against both jobs, reaching a verdict on each. . Worth 2 points. needs an explanation, not just an answer
7. The same verdict from two numbers that agree only once . 12 points. Question 7 of 10.

Completing the square on a quadratic leaves a squared binomial on one side and a constant on the other. That constant is not the discriminant $b^2 - 4ac$ , and it is worth finding out what it is instead, and what the two have to do with each other.
1. Part A.
  Solve $x^2 - 5x + 9 = 0$ without using the quadratic formula, giving both roots in the form $a + bi$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Now do the same to the general monic equation $x^2 + bx + c = 0$ : produce an equivalent equation whose left side is a single squared binomial and whose right side is one fraction.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  The right-hand side you produced in part B is a fraction, and the discriminant of $x^2 + bx + c$ is a whole number expression. For almost every $b$ and $c$ they are different numbers. Explain why they nevertheless always sort a monic quadratic into the same one of the three cases, and be precise about what is really being compared: the two numbers, or something less than the two numbers.
  
  Carry your own answer forward Compare whichever right-hand side you produced in part B against the discriminant, even if your part B was not the expected one, and say honestly whether they agree and why. The credit is for identifying what is actually being compared and for the argument about what preserves it.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer
Part A

$x = \dfrac{5}{2} + \dfrac{\sqrt{11}}{2}i$ and $x = \dfrac{5}{2} - \dfrac{\sqrt{11}}{2}i$ .
- $x = \frac{5 \pm i\sqrt{11}}{2}$ is the same pair before the fraction is split into parts; the question asks for the $a + bi$ form, so the real and imaginary parts should be visible
Part B

$\left(x + \dfrac{b}{2}\right)^2 = \dfrac{b^2 - 4c}{4}$ .
- $\left(x + \frac{b}{2}\right)^2 = \frac{b^2}{4} - c$ is the same equation with the right side left as two pieces; the question asks for one fraction
Part C

The constant on the right is $\frac{b^2 - 4c}{4}$ , exactly one quarter of the discriminant. Dividing by the positive number $4$ can change a value but never a sign, and the three cases consult only the sign of the quantity being square-rooted. So the two always agree about the case and almost never about the number.
Worked solution

Part A

Half of $-5$ is $-\frac{5}{2}$ , so add $\frac{25}{4}$ to both sides; on the right, $-9 = -\frac{36}{4}$ .

$\left(x - \frac{5}{2}\right)^2 = -\frac{11}{4} \quad\Longrightarrow\quad x = \frac{5}{2} \pm \frac{\sqrt{11}}{2}i.$

The right side is negative, so the roots are a complex conjugate pair, using $\sqrt{-\frac{11}{4}} = \frac{i\sqrt{11}}{2}$ .

Part B

Every move is the numerical one in general letters. Move $c$ across and add $\frac{b^2}{4}$ to both sides.

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

The numerator $b^2 - 4c$ is the discriminant $b^2 - 4ac$ at $a = 1$ : it appears from a derivation that never mentioned the formula.

Part C

Part B's constant is $\frac{b^2 - 4c}{4}$ , exactly one quarter of the discriminant $b^2 - 4c$ . The three cases consult only the sign of the quantity under the root, never its value, and dividing by the positive number $4$ cannot change a sign:

$\frac{b^2 - 4c}{4} > 0 \iff b^2 - 4c > 0, \qquad \text{and likewise for } = 0 \text{ and } < 0.$

So the two agree on the case every time while disagreeing on the number in all but one instance. What is compared is the signs, not the values.

In one line

$x^2 - 5x + 9 = 0$ has the conjugate roots $x = \frac{5}{2} \pm \frac{\sqrt{11}}{2}i$ , and completing the square on $x^2 + bx + c = 0$ gives $\left(x + \frac{b}{2}\right)^2 = \frac{b^2 - 4c}{4}$ . That constant is exactly a quarter of the discriminant, and since only the SIGN of the quantity under a root is ever consulted, dividing by the positive number $4$ leaves every verdict unchanged.

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

Adds the square of half the linear coefficient to both sides and writes the left as a squared binomial containing that same halved coefficient. . Worth 2 points.

Takes the root of the negative right side with both signs, converting it to $i$ times a real root and splitting over the denominator. . Worth 2 points.

Reports both roots in the requested $a + bi$ form, real and imaginary parts separated. . Worth 1 point.

Part B 3 points

Carries the general letters through the same steps, adding the square of half the coefficient and factoring the left into a squared binomial. . Worth 2 points.

Combines the right-hand side into a single fraction over a common denominator. . Worth 1 point.

Part C 4 points

Identifies how the two quantities are related and makes that relationship show a sign cannot change, concluding for all three cases. . Worth 3 points. needs an explanation, not just an answer

States that the sign of each quantity, not its value, is what is compared, and why that is all the three cases ask for. . Worth 1 point.
8. A flare, and two questions the model answers in different ways . 12 points. Question 8 of 10.

A signal flare is fired straight up from the top of a $40$ -foot watchtower on the shore, leaving the launcher at $48$ feet per second. Its height in feet, $t$ seconds after firing, is modelled by

$h = -16t^2 + 48t + 40.$

Give exact answers, and add a decimal only where it helps you say what the answer means.
1. Part A.
  The flare falls past the tower and strikes the water at the tower's base, where the height is $0$ . Find when, reporting both roots of your equation before you decide anything.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  An observer claims the flare passed $90$ feet above the water. Settle the claim.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Each of the previous parts produced roots, and in each part you had to work out what those roots were telling you about the flare. Explain what your roots meant in part A and what they meant in part B, and say where the mathematics settled the matter on its own and where you had to bring in something it could not know.
  
  Carry your own answer forward Account for whichever roots your parts A and B produced, and read them against the flare honestly. The credit here is for distinguishing what the two parts' roots meant, not for having landed on particular numbers.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
The answer
Part A

The roots are $t = \dfrac{3 + \sqrt{19}}{2}$ and $t = \dfrac{3 - \sqrt{19}}{2}$ . The flare strikes the water at $t = \dfrac{3 + \sqrt{19}}{2} \approx 3.7$ seconds.
- $t = \frac{6 + 2\sqrt{19}}{4}$ is the same time unreduced; $t = \frac{3 \pm \sqrt{19}}{2}$ names both roots, only one of which is an answer
Part B

The claim is false. Setting the height to $90$ gives $8t^2 - 24t + 25 = 0$ , whose discriminant is $-224$ , so there is no real time at which the height is $90$ feet: the flare never gets that high.

Part C

Part A's root is a real time that the model computes and the situation forbids: the flare did not exist then, so rejecting it is MY decision, imported from the meaning of $t$ . Part B's roots are not times at all; no real solution exists, so the algebra itself delivered the verdict and there was nothing to decide.
Worked solution

Part A

Set $h = 0$ and divide by $-8$ to get $2t^2 - 6t - 5 = 0$ . With $c = -5$ the $-4ac$ term is positive.

$t = \frac{6 \pm \sqrt{36 + 40}}{4} = \frac{3 \pm \sqrt{19}}{2}.$

The root $\frac{3 - \sqrt{19}}{2} \approx -0.68$ falls before firing, so the flare strikes the water at $t = \frac{3 + \sqrt{19}}{2} \approx 3.7$ seconds.

Part B

Set $h = 90$ and reach standard form $8t^2 - 24t + 25 = 0$ . The discriminant answers whether any real time exists.

$b^2 - 4ac = 576 - 800 = -224 < 0.$

It is negative, so the roots are complex, and a complex number is not a time. The flare never reaches $90$ feet, so the observer is mistaken.

Part C

The two rejections come from opposite directions. In part A, $\frac{3 - \sqrt{19}}{2} \approx -0.68$ is a genuine real solution of the equation; what rules it out is that $t$ is time from firing, and the flare had not been fired, so rejecting it is my decision, from the meaning of $t$ .

$\text{A: two real roots, one outside the situation.} \qquad \text{B: no real roots at all.}$

In part B the discriminant $-224$ makes the roots complex, and a complex number cannot be a time, so nothing was rejected: the algebra itself returned the verdict.

In one line

The flare strikes the water at $t = \frac{3 + \sqrt{19}}{2} \approx 3.7$ seconds, the root $\frac{3 - \sqrt{19}}{2} \approx -0.68$ being a real time that lies before the flight began. The claim of $90$ feet is false: that height gives $8t^2 - 24t + 25 = 0$ with discriminant $-224$ , so no real time delivers it. The first rejection is a decision imported from the meaning of $t$ ; the second is a verdict the algebra returned on its own.

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

Sets the height to zero and solves correctly, handling the sign of the $-4ac$ term when the constant is negative. . Worth 2 points.

Reports both roots then rejects the inadmissible one with the reason named. . Worth 1 point.

States the surviving answer as a time, with units. . Worth 1 point.

Part B 4 points

Sets the model equal to the stated height, reaches standard form, and computes the discriminant. . Worth 2 points.

Reads the discriminant back into the situation and answers the observer's claim directly. . Worth 2 points.

Part C 4 points

Distinguishes the two rejections, giving each its own grounds rather than a shared reason. . Worth 3 points. needs an explanation, not just an answer

Attributes each verdict to its source: one from the meaning imposed on the variable, one from the mathematics. . Worth 1 point.
9. A coefficient that is free to move, and one value it must not take . 12 points. Question 9 of 10.

Consider the equation

$kx^2 - 8x + 2 = 0,$

one equation for each real number $k$ . Only the leading coefficient moves; the other two never change.
1. Part A.
  Find every value of $k$ for which the equation has a repeated root, and give that root.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Describe the rest of the family. For which values of $k$ does the equation have two distinct real roots, and for which does it have a complex conjugate pair?
  
  Carry your own answer forward Use whichever discriminant and boundary value you produced in part A. Describe the two regions your own boundary separates, and say which side is which.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
3. Part C.
  There is one real number that belongs in none of the three cases you have described, and the discriminant has a confident opinion about it that should be ignored. Identify the value, say what the discriminant claims happens there and what actually happens, and explain why the claim carries no authority.
  
  Carry your own answer forward This part is about the structure of the family and not about your numbers, so give the argument in full even if parts A and B did not come out. Name the value your own answers should have excluded and say why.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

$k = 8$ , and the repeated root is $x = \dfrac{1}{2}$ .

Part B

Two distinct real roots when $k < 8$ (excluding $k = 0$ , which gives no quadratic at all); a complex conjugate pair when $k > 8$ . The value $k = 8$ found in part A is the boundary between the two.

Part C

At $k = 0$ the discriminant is $64 > 0$ and claims two distinct real roots. In fact $-8x + 2 = 0$ is linear and has the single root $x = \frac{1}{4}$ . The claim carries no authority because the three-case rule is derived by dividing by $a$ , so it presupposes $a \ne 0$ : fed a non-quadratic, it is not wrong so much as inapplicable.

Worked solution

Part A

A repeated root is the case where the discriminant vanishes. With $a = k$ , $b = -8$ , $c = 2$ :

$b^2 - 4ac = 64 - 8k = 0 \quad\Longrightarrow\quad k = 8.$

Then the formula collapses to $x = \frac{-b}{2a} = \frac{8}{16} = \frac{1}{2}$ , and indeed $8x^2 - 8x + 2 = 2(2x - 1)^2$ .

Part B

Only the sign of $64 - 8k$ matters. It is positive when

$64 - 8k > 0 \iff k < 8,$

giving two distinct real roots, and negative when $k > 8$ , giving a complex pair. At $k = 0$ there is no $x^2$ term, so exclude it: the real-root case is $k < 8$ with $k \ne 0$ .

Part C

The value is $k = 0$ . Then the $x^2$ term is gone and the equation is linear:

$-8x + 2 = 0 \quad\Longrightarrow\quad x = \frac{1}{4},$

a single root. Fed $k = 0$ , the discriminant returns $64$ and claims two real roots. It carries no authority because completing the square on $ax^2 + bx + c = 0$ divides by $a$ , so the whole three-case rule presupposes $a \ne 0$ . With $a = 0$ its hypothesis fails, so its output is not a claim about the equation at all.

In one line

$kx^2 - 8x + 2 = 0$ has discriminant $64 - 8k$ , so it has a repeated root only at $k = 8$ , where that root is $x = \frac{1}{2}$ ; two distinct real roots for $k < 8$ with $k \ne 0$ ; and a complex conjugate pair for $k > 8$ . At $k = 0$ the discriminant returns $64$ and claims two real roots, but the equation is the linear $-8x + 2 = 0$ with the single root $x = \frac{1}{4}$ : the three-case rule is derived by dividing by $a$ , so it says nothing when $a = 0$ .

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 3 points

Translates "a repeated root" into the discriminant being zero and writes it in terms of $k$ , with the unknown in the $4ac$ term. . Worth 2 points.

Solves for $k$ and produces the repeated root, by collapsing the formula or exhibiting a perfect square. . Worth 1 point.

Part B 4 points

Decides each case by the sign of the discriminant, an inequality in $k$ per case rather than sample values. . Worth 2 points.

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

Part C 5 points

Identifies the excluded value, states what the discriminant claims versus what the equation has, and traces the failure to a step in the formula's derivation. . Worth 3 points. needs an explanation, not just an answer

Says what the excluded case actually is, arguing from what the equation becomes there. . Worth 2 points. needs an explanation, not just an answer
10. Three sides of fence, and what the standard form knew in advance . 14 points. Question 10 of 10.

A rectangular exercise yard is to be built against a long straight wall, so fencing is needed on three sides only: two equal ends running out from the wall, and one side parallel to it. There are $44$ metres of fencing, all of it used, and the enclosed area must come to exactly $240$ square metres.
1. Part A.
  Name the unknown, write the other dimension in terms of it, and turn the area requirement into an equation in standard form with a leading coefficient of $1$ .
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 4 points
2. Part B.
  Solve your equation by rewriting one side as a squared binomial, and give both roots along with the dimensions each one produces.
  
  Carry your own answer forward Solve whichever equation you produced in part A, and build the dimensions from your own expression for the second side. The credit here is for the rewrite, for keeping both signs at the root, and for turning each root back into a pair of dimensions.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Decide which of your two roots the situation admits, and defend the decision by testing each one against everything the yard requires. Then show how your standard form from part A could have told you, before you solved anything, what the signs of its two roots had to be, and contrast this problem with an ordinary area problem in that respect.
  
  Carry your own answer forward Test whichever roots you found against the yard, and read your own standard form from part A rather than the expected one. The credit here is for putting each root to the situation honestly, for the argument you draw from the coefficients, and for the contrast, not for a particular pair of numbers.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer

Part A

With $x$ the length in metres of each end, the side parallel to the wall is $44 - 2x$ metres, and $x(44 - 2x) = 240$ becomes $x^2 - 22x + 120 = 0$ .
Part B

$(x - 11)^2 = 1$ , so $x = 10$ and $x = 12$ . The first gives a yard $10$ by $24$ metres and the second gives a yard $12$ by $20$ metres.
- $x = 11 \pm 1$ is the same pair; a $10$ by $24$ yard and a $12$ by $20$ yard are the two shapes, and reporting only the value of $x$ leaves the dimensions unstated
Part C

Both, since each gives positive dimensions and meets both requirements: two different yards fit the specification. The positive constant term makes the roots multiply to $+120$ , so they share a sign, and their sum of $+22$ makes that sign positive. A negative constant term is what splits the signs and forces a root out.
Worked solution

Part A

Let $x$ be each end, in metres. The wall takes no fencing, so the parallel side is $44 - 2x$ , and the area is the two dimensions multiplied.

$x(44 - 2x) = 240 \quad\Longrightarrow\quad x^2 - 22x + 120 = 0.$

Dividing through by the leading $-2$ makes the $x^2$ stand alone for the square.

Part B

Complete the square: half of $-22$ is $-11$ , $(-11)^2 = 121$ , and $-120 + 121 = 1$ .

$(x - 11)^2 = 1 \quad\Longrightarrow\quad x - 11 = \pm 1 \quad\Longrightarrow\quad x = 12 \text{ or } 10.$

With the parallel side $44 - 2x$ : at $x = 10$ the yard is $10$ by $24$ , and at $x = 12$ it is $12$ by $20$ .

Part C

Both roots are admitted. At $x = 10$ the sides are $10, 10, 24$ ; at $x = 12$ they are $12, 12, 20$ . Every length is positive, both use $44$ metres and enclose $240$ square metres, so two different yards fit.

The standard form knew this in advance. First check there are signs to read: the discriminant $484 - 480 = 4 > 0$ , so the roots are real. Then the sum and product, read off the coefficients,

$r_1 r_2 = 120, \qquad r_1 + r_2 = 22,$

show a positive product, so the roots share a sign, and a positive sum makes that sign positive. Both are positive, knowable from the coefficients alone. An ordinary area problem instead has a negative constant term, $w^2 + kw - A = 0$ , whose roots multiply to a negative number and so have opposite signs, forcing one out.

In one line

With $x$ the length of each end, the model is $x^2 - 22x + 120 = 0$ , which rewrites as $(x - 11)^2 = 1$ and gives $x = 10$ and $x = 12$ . Both are admitted: a $10$ by $24$ yard and a $12$ by $20$ yard each use $44$ metres of fencing and enclose $240$ square metres. The positive constant term forces the roots to multiply to $+120$ and so to share a sign, which is exactly why nothing had to be discarded here, unlike an area problem whose negative constant term splits the roots and rules one out.

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

Uses the fencing constraint to write the second dimension in one unknown, with the wall replacing one side. . Worth 2 points.

Turns the area requirement into a quadratic and divides every term by the leading coefficient to a monic standard form. . Worth 2 points.

Part B 4 points

Adds the square of half the linear coefficient to both sides and writes the left as a squared binomial containing that same halved value. . Worth 2 points.

Keeps both signs when taking the root and converts each root into the actual dimensions of a yard. . Worth 2 points.

Part C 6 points

Defends the decision by testing each root against every requirement, including the second dimension staying positive, rather than by reflex. . Worth 3 points. needs an explanation, not just an answer

Argues from the sign of the constant term, via the product of the roots, to what it forces about the roots, first establishing the roots have signs. . Worth 2 points. needs an explanation, not just an answer

Supplies the contrasting case, identifying the sign of the constant term as what differs and what it does to the roots. . Worth 1 point.

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. Two parts that keep their distance, and one operation that makes them meet 9 points

Part A 3

Combines like parts and distributes the minus sign across both parts of the second number in the difference. 2
Reports each result in standard form, one real part and one imaginary term. 1

Part B 3

Expands the product completely and resolves the $i^2$ term into a real contribution with the correct sign. 2
Collects the terms into standard form, one real part and one imaginary part. 1

Part C 3

Grounds the reason the parts cannot combine in $i$ not being a real number, rather than in a rule that is simply not allowed. 2
Locates the interaction in the product of two imaginary terms and names $i^2 = -1$ as what turns it real. 1

2. A square, a constant, and a verdict you can read without solving 11 points

Part A 3

Uses the square of half the coefficient of $x$ , adding and subtracting it on the same side rather than on a side that does not exist. 2
Reports a single squared binomial plus one combined constant, with the halved coefficient inside. 1

Part B 5

Isolates a squared binomial with a bare constant on the other side, adding the completing constant to both sides. 2
Takes the square root with both signs and converts the root of the negative into $i$ times a real square root. 2
Reports both roots in the requested $a + bi$ form, real and imaginary parts separated. 1

Part C 3

Argues from a real square never being negative to the whole expression never vanishing, not from the roots or the discriminant. 3

3. Two high powers, and a product worth looking at twice 11 points

Part A 3

Reduces each exponent using its remainder on division by $4$ and reports both powers with their signs. 2
Assembles the two results into standard form, real and imaginary values in their own parts. 1

Part B 4

Forms the conjugate by changing the sign of the imaginary part only, leaving the real part untouched. 1
Multiplies the pair out and resolves the $i^2$ correctly, leaving no imaginary part. 2
Notes out loud that the product has come out with no imaginary part. 1

Part C 4

Argues in general letters that the two middle terms are opposites that cancel, so no imaginary part survives for any complex number. 3
Shows the $i^2$ turns the subtracted square into an added one, and states which form the result takes. 1

4. One number, computed first, that names the answer before you have it 11 points

Part A 3

Reaches standard form before reading any coefficient, and reports all three coefficients with their signs. 2
Evaluates the discriminant and names the kind of roots it predicts, not just the number. 1

Part B 5

Substitutes into a correctly written formula, the whole numerator over $2a$ with the sign of $-b$ handled. 2
Handles the root of the discriminant, then reduces by a factor common to every term of the fraction. 2
Reports both roots in the requested $a + bi$ form, consistently with part A's verdict. 1

Part C 3

Locates the discriminant as the quantity the formula square-roots and ties the three cases to what a square root can do, rather than quoting them as a rule. 2
Says what the discriminant leaves undetermined: the values of the roots need information it does not carry. 1

5. A sail, and a number the algebra offers that the world will not take 11 points

Part A 3

Names the unknown and expresses the second dimension in that same unknown, no second letter. 2
Translates the area sentence into an equation and rearranges to standard form. 1

Part B 4

Solves the quadratic correctly, handling the sign of the $-4ac$ term when the constant is negative, and reports both roots first. 2
Tests each root against what the letter stands for and rejects the impossible one with the reason named. 1
Answers the question asked, giving both dimensions in a sentence with units. 1

Part C 4

Distinguishes satisfying the equation from answering the question, locating the rejection in the meaning of the letter rather than the algebra. 2
Reaches a verdict on the general rule and supports it with a specific situation, accounting for each of its roots. 2

6. Two quotients, and the two jobs one move has to do 12 points

Part A 4

Multiplies numerator and denominator by the conjugate of the denominator, produces a real denominator, and resolves the $i^2$ . 2
Divides the whole numerator to standard form and checks by multiplying back against the original numerator. 2

Part B 3

Treats the pure imaginary denominator as a complex number with real part zero and clears the $i$ from downstairs. 2
Resolves the $i^2$ , divides both parts, reduces, and reports standard form. 1

Part C 5

Separates the two jobs with the right reason for each: the multiplier is a fraction equal to $1$ so the value is preserved, and the conjugate product is real so the denominator is cleared. 3
Works out what the numerator's conjugate does against both jobs, reaching a verdict on each. 2

7. The same verdict from two numbers that agree only once 12 points

Part A 5

Adds the square of half the linear coefficient to both sides and writes the left as a squared binomial containing that same halved coefficient. 2
Takes the root of the negative right side with both signs, converting it to $i$ times a real root and splitting over the denominator. 2
Reports both roots in the requested $a + bi$ form, real and imaginary parts separated. 1

Part B 3

Carries the general letters through the same steps, adding the square of half the coefficient and factoring the left into a squared binomial. 2
Combines the right-hand side into a single fraction over a common denominator. 1

Part C 4

Identifies how the two quantities are related and makes that relationship show a sign cannot change, concluding for all three cases. 3
States that the sign of each quantity, not its value, is what is compared, and why that is all the three cases ask for. 1

8. A flare, and two questions the model answers in different ways 12 points

Part A 4

Sets the height to zero and solves correctly, handling the sign of the $-4ac$ term when the constant is negative. 2
Reports both roots then rejects the inadmissible one with the reason named. 1
States the surviving answer as a time, with units. 1

Part B 4

Sets the model equal to the stated height, reaches standard form, and computes the discriminant. 2
Reads the discriminant back into the situation and answers the observer's claim directly. 2

Part C 4

Distinguishes the two rejections, giving each its own grounds rather than a shared reason. 3
Attributes each verdict to its source: one from the meaning imposed on the variable, one from the mathematics. 1

9. A coefficient that is free to move, and one value it must not take 12 points

Part A 3

Translates "a repeated root" into the discriminant being zero and writes it in terms of $k$ , with the unknown in the $4ac$ term. 2
Solves for $k$ and produces the repeated root, by collapsing the formula or exhibiting a perfect square. 1

Part B 4

Decides each case by the sign of the discriminant, an inequality in $k$ per case rather than sample values. 2
Gets the direction of each inequality right, accounting for the sign with which $k$ enters. 2

Part C 5

Identifies the excluded value, states what the discriminant claims versus what the equation has, and traces the failure to a step in the formula's derivation. 3
Says what the excluded case actually is, arguing from what the equation becomes there. 2

10. Three sides of fence, and what the standard form knew in advance 14 points

Part A 4

Uses the fencing constraint to write the second dimension in one unknown, with the wall replacing one side. 2
Turns the area requirement into a quadratic and divides every term by the leading coefficient to a monic standard form. 2

Part B 4

Adds the square of half the linear coefficient to both sides and writes the left as a squared binomial containing that same halved value. 2
Keeps both signs when taking the root and converts each root into the actual dimensions of a yard. 2

Part C 6

Defends the decision by testing each root against every requirement, including the second dimension staying positive, rather than by reflex. 3
Argues from the sign of the constant term, via the product of the roots, to what it forces about the roots, first establishing the roots have signs. 2
Supplies the contrasting case, identifying the sign of the constant term as what differs and what it does to the roots. 1

Multiple choice

Free response