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Special Manipulations: 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, 92 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 numbers known only by their sum and product . 9 points. Question 1 of 10.

Two numbers $x$ and $y$ have sum $x + y = 7$ and product $xy = 4$ . Neither number is given on its own, and neither is needed.
1. Part A.
  Find $x^2 + y^2$ and $\dfrac{1}{x} + \dfrac{1}{y}$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Find $x^3 + y^3$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Explain why $x^2 + y^2$ reduces to $s^2 - 2p$ and not $s^2 + 2p$ .
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$x^2 + y^2 = 41$ and $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{7}{4}$ .

Part B

$x^3 + y^3 = 259$ .

Part C

Because $(x + y)^2 = x^2 + 2xy + y^2$ , the square of the sum already contains the cross term $2xy = 2p$ . Isolating $x^2 + y^2$ means removing that term, so it is subtracted; adding $2p$ would count the cross term a second time.

Worked solution

Part A

Read off $s = 7$ and $p = 4$ . The sum of squares subtracts the cross term, and the reciprocal sum is the sum over the product.

$x^2 + y^2 = s^2 - 2p = 49 - 8 = 41, \qquad \frac{1}{x} + \frac{1}{y} = \frac{s}{p} = \frac{7}{4}$

Part B

Cube the sum; the correction term carries a factor of $s$ .

$x^3 + y^3 = s^3 - 3ps = 343 - 3(4)(7) = 343 - 84 = 259$

Part C

Expand the square of the sum.

$(x + y)^2 = x^2 + 2xy + y^2 = s^2$

The middle term $2xy$ equals $2p$ . To recover $x^2 + y^2$ alone, subtract it from $s^2$ , giving $s^2 - 2p$ . Writing $s^2 + 2p$ would add the cross term that is already inside $s^2$ , double-counting it.

In one line

With $s = 7$ and $p = 4$ : $x^2 + y^2 = 41$ , $\frac{1}{x} + \frac{1}{y} = \frac{7}{4}$ , and $x^3 + y^3 = 259$ . The sum of squares subtracts $2p$ because $(x + y)^2$ already carries the cross term $2xy = 2p$ , which must be removed to leave $x^2 + y^2$ .

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

Applies $s^2 - 2p$ and $\tfrac{s}{p}$ to the given sum and product. . Worth 2 points.

Reports both values, obtained from the symmetric totals without solving for $x$ and $y$ separately. . Worth 1 point.

Part B 3 points

Uses $s^3 - 3ps$ with the factor of $s$ kept in the correction term. . Worth 2 points.

Reports the value as the sum of cubes of the pair. . Worth 1 point.

Part C 3 points

Derives from $(x + y)^2 = x^2 + 2xy + y^2$ that the cross term $2p$ sits inside $s^2$ and must be subtracted. . Worth 3 points. needs an explanation, not just an answer
2. A radical equation and the candidates it produces . 9 points. Question 2 of 10.

Consider the equation $\sqrt{x + 7} = x - 5$ .
1. Part A.
  Square both sides, solve the resulting quadratic, and list both candidates for $x$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Test each candidate in the original equation and state which are genuine solutions.
  
  Carry your own answer forward Test whichever candidates you produced in part A.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Explain why squaring both sides can introduce a candidate that fails the original, and what condition on $x$ the equation $\sqrt{x + 7} = x - 5$ imposes on its own.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

Squaring gives $x^2 - 11x + 18 = 0$ , so the candidates are $x = 2$ and $x = 9$ .

Part B

Only $x = 9$ : it gives $\sqrt{16} = 4 = 9 - 5$ . The candidate $x = 2$ gives $\sqrt{9} = 3 \neq 2 - 5$ , so it is extraneous.

Part C

Squaring sends $a$ and $-a$ to the same value, so the squared equation also holds where $\sqrt{x + 7} = -(x - 5)$ , a branch the original never had. And a principal root is nonnegative, so the original forces $x - 5 \geq 0$ , that is $x \geq 5$ , which $x = 2$ breaks.

Worked solution

Part A

Square both sides, expanding the binomial in full, then gather into a quadratic and factor.

$x + 7 = (x - 5)^2 = x^2 - 10x + 25 \ \Longrightarrow\ x^2 - 11x + 18 = 0 \ \Longrightarrow\ (x - 2)(x - 9) = 0$

Part B

Substitute each candidate into the original $\sqrt{x + 7} = x - 5$ , whose right side cannot be negative.

$x = 9\!:\ \sqrt{16} = 4 = 9 - 5\ \checkmark \qquad x = 2\!:\ \sqrt{9} = 3 \neq 2 - 5 = -3$

Part C

Squaring is not reversible: from $a^2 = b^2$ one gets $a = b$ or $a = -b$ , so the squared equation carries the extra branch $\sqrt{x + 7} = -(x - 5)$ .

$\sqrt{x + 7} \geq 0 \ \Longrightarrow\ x - 5 \geq 0 \ \Longrightarrow\ x \geq 5$

The candidate $x = 2$ solves the extra branch but breaks $x \geq 5$ , which is why the check in the original rejects it.

In one line

Squaring $\sqrt{x + 7} = x - 5$ gives $x^2 - 11x + 18 = 0$ with candidates $x = 2$ and $x = 9$ ; only $x = 9$ checks, and $x = 2$ is extraneous because the original demands $x - 5 \geq 0$ , that is $x \geq 5$ .

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

Squares both sides, expanding the binomial in full, and forms the quadratic. . Worth 2 points.

Lists both roots as candidates, not yet confirmed solutions. . Worth 1 point.

Part B 3 points

Substitutes each candidate into the original equation, not the squared one. . Worth 2 points.

Keeps only the candidate that checks, discarding the extraneous one. . Worth 1 point.

Part C 3 points

Traces the extra candidate to the second branch squaring admits, and names the nonnegativity condition $x - 5 \geq 0$ the original imposes. . Worth 3 points. needs an explanation, not just an answer
3. A fraction that contains a copy of itself . 9 points. Question 3 of 10.

The continued fraction $x = 6 + \cfrac{1}{6 + \cfrac{1}{6 + \cdots}}$ repeats the same pattern below every division bar.
1. Part A.
  Name the fraction $x$ , write the finite equation it must satisfy, and clear it to a quadratic.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Solve the quadratic and give the value of the continued fraction.
  
  Carry your own answer forward Solve whichever quadratic you obtained in part A.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  The value satisfies $x = 6 + \dfrac{1}{x}$ . Interpret what this identity says about how the number compares with its own reciprocal.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points
The answer

Part A

$x = 6 + \dfrac{1}{x}$ , which clears to $x^2 - 6x - 1 = 0$ .

Part B

$x = 3 + \sqrt{10}$ ; the root $3 - \sqrt{10}$ is negative and is rejected.

Part C

Rearranged, $x - \dfrac{1}{x} = 6$ , so the number exceeds its own reciprocal by exactly $6$ . That gap of $6$ is the leading $6 +$ of the fraction, so the self-similarity is just this relationship between $x$ and $\dfrac{1}{x}$ restated.

Worked solution

Part A

Below the first $6 +$ sits $1$ over a perfect copy of the whole fraction, again $x$ , so the tail is $\dfrac{1}{x}$ . Multiply through by $x$ .

$x = 6 + \frac{1}{x} \ \Longrightarrow\ x^2 = 6x + 1 \ \Longrightarrow\ x^2 - 6x - 1 = 0$

Part B

Apply the quadratic formula, then keep the positive root, since a fraction built from positive parts is positive.

$x = \frac{6 \pm \sqrt{36 + 4}}{2} = 3 \pm \sqrt{10} \ \Longrightarrow\ x = 3 + \sqrt{10}$

Part C

Subtract $\dfrac{1}{x}$ from both sides of the self-similar identity.

$x - \frac{1}{x} = 6$

So $x$ is exactly $6$ more than its reciprocal. The constant $6$ that opens each layer of the fraction is precisely that gap, which is why the endless fraction and this one-line relationship carry the same information.

In one line

Naming the fraction $x$ gives $x = 6 + \frac{1}{x}$ , so $x^2 - 6x - 1 = 0$ and $x = 3 + \sqrt{10}$ (the negative root is rejected). The identity $x = 6 + \frac{1}{x}$ says $x$ exceeds its reciprocal by exactly $6$ , which is the fraction's opening $6$ restated.

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

Identifies the tail below the first $6 +$ as the copy $\tfrac{1}{x}$ and writes $x = 6 + \tfrac{1}{x}$ . . Worth 2 points.

Clears the fraction to a quadratic in $x$ . . Worth 1 point.

Part B 3 points

Solves the quadratic by the formula or by completing the square. . Worth 2 points.

Selects the positive root as the value, rejecting the negative one. . Worth 1 point.

Part C 3 points

Reads $x = 6 + \tfrac{1}{x}$ as saying the number exceeds its reciprocal by exactly $6$ , tying the gap to the fraction's leading term. . Worth 3 points. needs an explanation, not just an answer
4. Undoing even powers and the sign they hide . 9 points. Question 4 of 10.

This question is about how raising to a power interacts with the sign of a solution.
1. Part A.
  Solve $(x - 3)^2 = 16$ for all real $x$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Solve $x^{2/3} = 9$ for all real $x$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Explain why undoing an even power can require keeping both signs while undoing an odd one does not, using how $x^2$ and $x^3$ treat the sign of $x$ .
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

$x = 7$ or $x = -1$ .

Part B

$x = 27$ or $x = -27$ .

Part C

$x^2$ sends $x$ and $-x$ to the same value, so undoing it cannot tell which sign was there and both must be kept, giving $\pm$ . $x^3$ sends $x$ and $-x$ to different values, so it is reversible and a single sign is recovered, with no $\pm$ .

Worked solution

Part A

Take the square root of both sides; the absolute value forces two signs.

$\lvert x - 3 \rvert = 4 \ \Longrightarrow\ x - 3 = \pm 4 \ \Longrightarrow\ x = 7 \text{ or } x = -1$

Part B

Read $x^{2/3} = (x^{1/3})^2$ and let $u = x^{1/3}$ . The even power gives two signs, and cubing each recovers $x$ .

$u^2 = 9 \ \Longrightarrow\ u = \pm 3 \ \Longrightarrow\ x = (\pm 3)^3 = \pm 27$

Part C

Compare the two maps on a number and its opposite.

$(-x)^2 = x^2 \qquad\text{but}\qquad (-x)^3 = -x^3$

Squaring collapses $x$ and $-x$ onto one value, so reversing it restores both possibilities and a $\pm$ is unavoidable. Cubing keeps them distinct, so reversing it names one sign, and no $\pm$ appears.

In one line

$(x - 3)^2 = 16$ gives $x = 7$ or $x = -1$ , and $x^{2/3} = 9$ gives $x = \pm 27$ . Both carry two signs because an even power sends $x$ and $-x$ to the same value, so undoing it must keep both; an odd power keeps them distinct and needs no $\pm$ .

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

Takes the root of both sides and keeps the $\pm$ from the even root. . Worth 2 points.

Reports both solutions, not only the positive branch. . Worth 1 point.

Part B 3 points

Sets $u = x^{1/3}$ , finds $u = \pm 3$ , and cubes to recover $x$ . . Worth 2 points.

Reports both values, the even numerator producing two. . Worth 1 point.

Part C 3 points

Explains that an even power identifies $x$ with $-x$ , so both signs survive undoing it, while an odd power keeps them distinct. . Worth 3 points. needs an explanation, not just an answer
5. A tower of square roots, named and checked . 9 points. Question 5 of 10.

The nested radical $\sqrt{56 + \sqrt{56 + \sqrt{56 + \cdots}}}$ repeats itself under every radical sign.
1. Part A.
  Name the tower $x$ , write the finite equation it satisfies, and turn it into a quadratic.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Solve the quadratic, then say which root is the value of the tower and why the other is discarded.
  
  Carry your own answer forward Use the quadratic you formed in part A.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  The squared equation had two roots, but the tower has one value. Explain why squaring the self-similar equation is the same source of an extra candidate as squaring a radical equation, and what rejects the extra one here.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$x = \sqrt{56 + x}$ , so $x^2 - x - 56 = 0$ .

Part B

$(x - 8)(x + 7) = 0$ , so $x = 8$ or $x = -7$ ; the value is $x = 8$ , since a tower of principal square roots cannot be negative, so $-7$ is discarded.

Part C

Squaring is even, so $x = \sqrt{56 + x}$ and $x = -\sqrt{56 + x}$ share the squared equation $x^2 = 56 + x$ ; the negative branch is the extra candidate. It is rejected exactly like an extraneous root: the value must satisfy the original $x = \sqrt{56 + x}$ , whose right side is nonnegative, so $-7$ cannot be it.

Worked solution

Part A

Under the outermost radical sits $56$ plus a perfect copy of the whole tower, which is $x$ . Square to clear the root.

$x = \sqrt{56 + x} \ \Longrightarrow\ x^2 = 56 + x \ \Longrightarrow\ x^2 - x - 56 = 0$

Part B

Factor and take both roots, then keep the nonnegative one.

$(x - 8)(x + 7) = 0 \ \Longrightarrow\ x = 8 \text{ or } x = -7$

A nest of principal square roots of positive quantities is nonnegative, so the value is $x = 8$ and $-7$ is discarded.

Part C

The step $x^2 = 56 + x$ is satisfied by both $x = \sqrt{56 + x}$ and $x = -\sqrt{56 + x}$ , because squaring erases the sign, exactly as when solving a radical equation.

$x = \sqrt{56 + x} \geq 0$

The original self-similar equation demands a nonnegative left side, so the negative branch, and with it the root $-7$ , is extraneous and discarded.

In one line

Naming the tower $x$ gives $x = \sqrt{56 + x}$ , so $x^2 - x - 56 = 0$ with roots $8$ and $-7$ ; the value is $x = 8$ . Squaring admitted the branch $x = -\sqrt{56 + x}$ , whose root $-7$ is extraneous because the original tower is nonnegative, exactly as an extraneous root is rejected in a radical equation.

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

Replaces the inner copy by $x$ and writes $x = \sqrt{56 + x}$ . . Worth 2 points.

Squares to reach a quadratic in $x$ . . Worth 1 point.

Part B 3 points

Solves the quadratic to both roots. . Worth 2 points.

Selects the nonnegative root as the value and discards the negative one. . Worth 1 point.

Part C 3 points

Attributes the extra root to squaring admitting the negative branch, and rejects it by the nonnegativity of the original tower, as with an extraneous root. . Worth 3 points. needs an explanation, not just an answer
6. When an infinite sum has a value and when it has none . 9 points. Question 6 of 10.

Three infinite geometric series are given: $P = 4 + 2 + 1 + \tfrac{1}{2} + \cdots$ , $Q = 4 + 8 + 16 + 32 + \cdots$ , and $R = 4 - 2 + 1 - \tfrac{1}{2} + \cdots$ .
1. Part A.
  For each series, state its common ratio and whether it settles on a finite value.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  For each series that settles, find its sum.
  
  Carry your own answer forward Sum whichever series you judged to settle in part A.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  The formula $\dfrac{a_1}{1 - r}$ can be evaluated for every ratio, including $r = 2$ . Explain why it gives the true sum only when $\lvert r \rvert < 1$ , by reference to the derivation $S = a_1 + rS$ and to what the running totals do when $\lvert r \rvert \geq 1$ .
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$P$ : $r = \tfrac{1}{2}$ , settles; $Q$ : $r = 2$ , does not settle; $R$ : $r = -\tfrac{1}{2}$ , settles.

Part B

$P = \dfrac{4}{1 - 1/2} = 8$ and $R = \dfrac{4}{1 - (-1/2)} = \dfrac{8}{3}$ ; $Q$ has no sum.

Part C

The formula comes from $S = a_1 + rS$ , which assumes $S$ exists. When $\lvert r \rvert < 1$ the terms shrink and the running totals close in on a number, so the assumption holds. When $\lvert r \rvert \geq 1$ the terms do not shrink to zero and the totals run off or oscillate, so no $S$ exists and the output is meaningless.

Worked solution

Part A

Divide consecutive terms to get each ratio, then apply $\lvert r \rvert < 1$ .

$P\!:\ r = \tfrac{1}{2}, \quad Q\!:\ r = 2, \quad R\!:\ r = -\tfrac{1}{2}$

Only $P$ and $R$ have $\lvert r \rvert < 1$ , so only they settle; $Q$ grows.

Part B

Use $\dfrac{a_1}{1 - r}$ for the settling series, keeping the sign of $r$ .

$P = \frac{4}{1 - \tfrac{1}{2}} = 8, \qquad R = \frac{4}{1 - (-\tfrac{1}{2})} = \frac{4}{\tfrac{3}{2}} = \frac{8}{3}$

Part C

The closed form is derived by writing $S = a_1 + rS$ and solving, a step that already assumes there is a number $S$ .

$S = a_1 + rS \ \Longrightarrow\ S = \frac{a_1}{1 - r}$

When $\lvert r \rvert < 1$ the terms shrink to zero and the partial sums close in on one value, so the assumption is sound. When $\lvert r \rvert \geq 1$ the terms do not vanish and the partial sums grow or oscillate forever, so no $S$ exists and the number the formula prints is the value of an empty assumption.

In one line

The ratios are $r = \tfrac{1}{2}$ for $P$ , $r = 2$ for $Q$ , and $r = -\tfrac{1}{2}$ for $R$ , so $P$ and $R$ settle while $Q$ does not: $P = 8$ and $R = \tfrac{8}{3}$ . The formula $\tfrac{a_1}{1 - r}$ gives a true sum only when $\lvert r \rvert < 1$ , because its derivation $S = a_1 + rS$ assumes a sum exists, which fails once the terms no longer shrink to 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

Finds the common ratio of each series by dividing consecutive terms. . Worth 2 points.

Applies $\lvert r \rvert < 1$ to decide which series settle. . Worth 1 point.

Part B 3 points

Applies $\tfrac{a_1}{1 - r}$ with the sign of $r$ handled for each settling series. . Worth 2 points.

Leaves the non-settling series unsummed rather than forcing a value. . Worth 1 point.

Part C 3 points

Ties the formula to the assumption $S = a_1 + rS$ and explains that only $\lvert r \rvert < 1$ makes the partial sums settle, so only then is the output a true sum. . Worth 3 points. needs an explanation, not just an answer
7. A system that hides its product . 9 points. Question 7 of 10.

Two real numbers satisfy $x + y = 8$ and $x^2 + y^2 = 40$ .
1. Part A.
  Find the product $xy$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Using the sum and the product, find the pair $\{x, y\}$ .
  
  Carry your own answer forward Use the sum from the stem and the product you found in part A.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  The system fixes the pair but not which number is $x$ . Explain why, and state what feature of $s$ and $p$ guarantees the two numbers are real and distinct.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points
The answer

Part A

$xy = 12$ .

Part B

$\{x, y\} = \{2, 6\}$ .

Part C

Swapping $x$ and $y$ leaves both equations unchanged, so the system cannot tell them apart; both $(2, 6)$ and $(6, 2)$ solve it, so only the unordered pair is fixed. They are real and distinct because $(x - y)^2 = s^2 - 4p = 64 - 48 = 16 > 0$ .

Worked solution

Part A

The sum is $s = 8$ , but the product is not given. Recover it from $x^2 + y^2 = s^2 - 2p$ .

$40 = 8^2 - 2p \ \Longrightarrow\ 2p = 64 - 40 = 24 \ \Longrightarrow\ p = 12$

Part B

The two numbers are the roots of $t^2 - st + p = 0$ with $s = 8$ and $p = 12$ .

$t^2 - 8t + 12 = 0 \ \Longrightarrow\ (t - 2)(t - 6) = 0 \ \Longrightarrow\ \{x, y\} = \{2, 6\}$

Part C

Interchanging $x$ and $y$ turns $x + y = 8$ into $y + x = 8$ and $x^2 + y^2 = 40$ into $y^2 + x^2 = 40$ , the same two equations, so no solution can single out one variable.

$(x - y)^2 = s^2 - 4p = 64 - 48 = 16 > 0$

A positive value for $(x - y)^2$ means $x - y$ is a nonzero real number, so the two are real and unequal.

In one line

From $x + y = 8$ and $x^2 + y^2 = 40$ , the product is $p = 12$ , so $t^2 - 8t + 12 = 0$ and $\{x, y\} = \{2, 6\}$ . The system is symmetric, so it fixes only the unordered pair, and $(x - y)^2 = s^2 - 4p = 16 > 0$ guarantees the pair is real and distinct.

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

Rearranges $x^2 + y^2 = s^2 - 2p$ to solve for $p$ . . Worth 2 points.

Identifies the result as the product $xy$ . . Worth 1 point.

Part B 3 points

Forms $t^2 - st + p = 0$ from the sum and product and solves it. . Worth 2 points.

Reports the result as the unordered pair. . Worth 1 point.

Part C 3 points

Explains that the swap-invariance of the system yields only the unordered pair, and reads $s^2 - 4p > 0$ as the guarantee of a real, distinct pair. . Worth 3 points. needs an explanation, not just an answer
8. Two manipulations that change the solution set . 9 points. Question 8 of 10.

Two equations are handled by an unbalanced move. Equation I is $\dfrac{x^2}{x - 3} = \dfrac{9}{x - 3}$ , and Equation II is $x^2 = 6x$ .
1. Part A.
  Solve Equation I. Multiplying both sides by $x - 3$ gives $x^2 = 9$ ; decide which of its roots are genuine solutions of the original.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Solve Equation II correctly, and identify the solution that dividing both sides by $x$ would discard.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Explain why multiplying both sides by an expression that can be zero can ADD a solution, while dividing by one can LOSE a solution.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

$x = -3$ only. Multiplying gives $x = \pm 3$ , but $x = 3$ makes the original denominators zero, so it is outside the domain and rejected.

Part B

$x = 0$ or $x = 6$ ; dividing both sides by $x$ would discard $x = 0$ .

Part C

Multiplying by $x - 3$ makes both sides equal, both zero, wherever $x - 3 = 0$ , so it manufactures a solution at $x = 3$ the original need not have. Dividing by $x$ assumes $x \neq 0$ and silently drops any solution where $x = 0$ , because that case is thrown away before solving.

Worked solution

Part A

Multiplying both sides by $x - 3$ clears the denominators.

$x^2 = 9 \ \Longrightarrow\ x = \pm 3$

But $x = 3$ makes the original denominators zero, where the equation is undefined, so it was introduced by multiplying by the vanishing factor $x - 3$ . Only $x = -3$ survives.

Part B

Bring everything to one side and factor rather than dividing by $x$ .

$x^2 - 6x = 0 \ \Longrightarrow\ x(x - 6) = 0 \ \Longrightarrow\ x = 0 \text{ or } x = 6$

Dividing by $x$ assumes $x \neq 0$ and throws away the root $x = 0$ .

Part C

A multiplication or a division by a variable expression is not reversible where that expression vanishes.

$x - 3 = 0 \ \Rightarrow\ \text{both sides become } 0; \qquad \text{dividing by } x \ \Rightarrow\ \text{assumes } x \neq 0$

Multiplying by $x - 3$ forces equality at $x = 3$ whether or not the original held there, adding a candidate. Dividing by $x$ discards $x = 0$ before the solving even begins, losing a genuine root.

In one line

Equation I gives $x^2 = 9$ , but $x = 3$ is rejected as it zeroes the original denominators, leaving $x = -3$ ; Equation II factors to $x(x - 6) = 0$ , so $x = 0$ or $x = 6$ , and dividing by $x$ would lose $x = 0$ . Multiplying by a vanishing factor forces a false equality and adds a root, while dividing by one drops the case where it is zero and loses a root.

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

Solves $x^2 = 9$ to $x = \pm 3$ , keeping both signs. . Worth 2 points.

Discards $x = 3$ as outside the original's domain, keeping only $x = -3$ . . Worth 1 point.

Part B 3 points

Factors $x(x - 6) = 0$ instead of dividing by $x$ . . Worth 2 points.

Names $x = 0$ as the solution that dividing by $x$ would lose. . Worth 1 point.

Part C 3 points

Explains that a vanishing factor forces equality at its zero (adding a solution) while dividing by it drops the zero case (losing a solution). . Worth 3 points. needs an explanation, not just an answer
9. A self-similar computation to audit . 9 points. Question 9 of 10.

Here is a proposed evaluation of $S = 1 + 6 + 36 + 216 + \cdots$ , given line by line.

Line 1: factor $6$ from every term after the first, so $S = 1 + 6(1 + 6 + 36 + \cdots)$ .

Line 2: the bracket is a copy of $S$ , so $S = 1 + 6S$ .

Line 3: solve, giving $-5S = 1$ , hence $S = -\tfrac{1}{5}$ .
1. Part A.
  Every line above is a valid algebraic manipulation, yet the conclusion $S = -\tfrac{1}{5}$ is nonsense. Identify the assumption that makes the whole computation invalid, and show how the partial sums of the series expose it.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 3 points
2. Part B.
  Contrast this with $T = 1 + \tfrac{1}{6} + \tfrac{1}{36} + \cdots$ . Does the same self-similar move give a trustworthy value here, and if so, what is it?
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  State the general condition that separates the two cases, and explain why the identical algebra is trustworthy for one series and empty for the other.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

The computation assumes from its first symbol that the series has a sum $S$ . It does not: with ratio $6$ the partial sums $1, 7, 43, 259, \ldots$ grow without bound and never settle, so there is no number $S$ , and every line drawn from ' $S =$ ' is empty.

Part B

Yes: $r = \tfrac{1}{6}$ with $\lvert r \rvert < 1$ , so the move is valid and $T = \dfrac{1}{1 - 1/6} = \dfrac{6}{5}$ .

Part C

For a geometric series the self-similar value is trustworthy exactly when $\lvert r \rvert < 1$ : only then do the terms shrink and the partial sums close in on a number. With $r = 6$ the sums diverge, so the assumption behind the algebra is false; with $r = \tfrac{1}{6}$ they converge, so it is sound and the same steps give a real value.

Worked solution

Part A

No single line is a false algebra step; the fault is the premise. Writing $S = \ldots$ presumes the series names a number.

$1,\ 7,\ 43,\ 259,\ 1555,\ \ldots \longrightarrow \infty$

The running totals march off without approaching anything, so no number $S$ exists. Because the premise is empty, the flawless algebra that follows, including $S = -\tfrac{1}{5}$ , carries no meaning.

Part B

Apply the same move: $T = 1 + \tfrac{1}{6}T$ . The ratio is $\tfrac{1}{6}$ , and $\lvert \tfrac{1}{6} \rvert < 1$ , so the series settles and the value is trustworthy.

$T - \tfrac{1}{6}T = 1 \ \Longrightarrow\ \tfrac{5}{6}T = 1 \ \Longrightarrow\ T = \tfrac{6}{5}$

Part C

The algebra is identical in both cases, so what differs is only whether the series has a sum to find.

$\lvert r \rvert < 1 \ \Longrightarrow\ \text{partial sums settle}; \qquad \lvert r \rvert \geq 1 \ \Longrightarrow\ \text{they do not}$

With $r = 6 \geq 1$ the assumption that $S$ exists is false, so the move is empty; with $r = \tfrac{1}{6} < 1$ it is true, so the move is trustworthy. The condition $\lvert r \rvert < 1$ is the whole difference.

In one line

The computation is invalid because it assumes $S = 1 + 6 + 36 + \cdots$ is a number, but with ratio $6$ the partial sums $1, 7, 43, 259, \ldots$ diverge, so no $S$ exists. The same move on $T = 1 + \tfrac{1}{6} + \cdots$ is trustworthy, giving $T = \tfrac{6}{5}$ , because $r = \tfrac{1}{6}$ satisfies $\lvert r \rvert < 1$ : that condition is the whole difference.

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

Locates the fault in the premise that the series has a sum, rather than in any single algebra step. . Worth 1 point.

Shows the partial sums grow without bound, so no number $S$ exists to name. . Worth 2 points. needs an explanation, not just an answer

Part B 3 points

Applies the self-similar move and notes $\lvert r \rvert < 1$ makes it valid. . Worth 2 points.

Reports $T = \tfrac{3}{2}$ as a genuine value, not an empty one. . Worth 1 point.

Part C 3 points

States the $\lvert r \rvert < 1$ condition and ties the trustworthiness of the identical algebra to whether the partial sums settle. . Worth 3 points. needs an explanation, not just an answer
10. An equation whose coefficients read the same both ways . 11 points. Question 10 of 10.

The equation $6x^4 - 35x^3 + 62x^2 - 35x + 6 = 0$ has coefficients $6, -35, 62, -35, 6$ , a palindrome.
1. Part A.
  Confirm $x = 0$ is not a root, then divide by $x^2$ and substitute $u = x + \tfrac{1}{x}$ to reduce the quartic to a quadratic in $u$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Solve for $u$ , then back-substitute to find all four values of $x$ .
  
  Carry your own answer forward Solve whichever quadratic in $u$ you obtained in part A, then undo the substitution.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  The four roots split into two pairs, each pair multiplying to $1$ . Explain why every root $r$ of a palindromic equation is accompanied by $\tfrac{1}{r}$ .
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$x = 0$ gives $6 \neq 0$ , so the division is safe; the reduction gives $6u^2 - 35u + 50 = 0$ .

Part B

$u = \tfrac{10}{3}$ or $u = \tfrac{5}{2}$ ; these give $x = 3, \tfrac{1}{3}, 2, \tfrac{1}{2}$ .

Part C

Reversing the coefficients of a degree- $n$ polynomial $P(x)$ produces $x^n P(\tfrac{1}{x})$ . A palindrome means $P(x) = x^n P(\tfrac{1}{x})$ , so $P(r) = 0$ forces $r^n P(\tfrac{1}{r}) = 0$ ; since $r \neq 0$ , $P(\tfrac{1}{r}) = 0$ , and $\tfrac{1}{r}$ is a root too.

Worked solution

Part A

Testing $x = 0$ gives the constant term $6 \neq 0$ , so $0$ is not a root and dividing by $x^2$ loses nothing. Divide, group the mirror terms, and substitute $x^2 + \tfrac{1}{x^2} = u^2 - 2$ .

$6\left(x^2 + \tfrac{1}{x^2}\right) - 35\left(x + \tfrac{1}{x}\right) + 62 = 6(u^2 - 2) - 35u + 62 = 6u^2 - 35u + 50 = 0$

Part B

Solve the quadratic in $u$ , then undo the substitution for each value.

$6u^2 - 35u + 50 = 0 \ \Longrightarrow\ u = \tfrac{10}{3} \text{ or } \tfrac{5}{2}$

For $u = \tfrac{10}{3}$ : $3x^2 - 10x + 3 = 0 \Rightarrow x = 3, \tfrac{1}{3}$ . For $u = \tfrac{5}{2}$ : $2x^2 - 5x + 2 = 0 \Rightarrow x = 2, \tfrac{1}{2}$ .

Part C

Reading a polynomial's coefficients backwards is the operation $P(x) \mapsto x^n P(\tfrac{1}{x})$ , which reverses the coefficient list. A palindrome is a polynomial equal to its own reversal.

$P(x) = x^n P\!\left(\tfrac{1}{x}\right)$

If $P(r) = 0$ then $r^n P(\tfrac{1}{r}) = P(r) = 0$ . Since $x = 0$ is not a root, $r \neq 0$ , so $r^n \neq 0$ and $P(\tfrac{1}{r}) = 0$ : the reciprocal $\tfrac{1}{r}$ is a root as well.

In one line

Dividing $6x^4 - 35x^3 + 62x^2 - 35x + 6 = 0$ by $x^2$ and substituting $u = x + \tfrac{1}{x}$ gives $6u^2 - 35u + 50 = 0$ , so $u = \tfrac{10}{3}$ or $\tfrac{5}{2}$ , and the roots are $x = 3, \tfrac{1}{3}, 2, \tfrac{1}{2}$ . They pair as reciprocals because a palindrome satisfies $P(x) = x^n P(\tfrac{1}{x})$ , so $P(r) = 0$ forces $P(\tfrac{1}{r}) = 0$ .

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

Checks that $x = 0$ is not a root, then divides by $x^2$ and groups the mirror terms. . Worth 2 points.

Substitutes $x^2 + \tfrac{1}{x^2} = u^2 - 2$ to reach a quadratic in $u$ . . Worth 1 point.

Part B 5 points

Solves the quadratic in $u$ for both values. . Worth 2 points.

Back-substitutes each $u$ , solving $x + \tfrac{1}{x} = u$ for $x$ . . Worth 2 points.

Reports all four values of $x$ . . Worth 1 point.

Part C 3 points

Uses the palindrome identity $P(x) = x^n P(\tfrac{1}{x})$ to show $P(r) = 0$ forces $P(\tfrac{1}{r}) = 0$ , given $r \neq 0$ . . Worth 3 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. Two numbers known only by their sum and product 9 points

Part A 3

Applies $s^2 - 2p$ and $\tfrac{s}{p}$ to the given sum and product. 2
Reports both values, obtained from the symmetric totals without solving for $x$ and $y$ separately. 1

Part B 3

Uses $s^3 - 3ps$ with the factor of $s$ kept in the correction term. 2
Reports the value as the sum of cubes of the pair. 1

Part C 3

Derives from $(x + y)^2 = x^2 + 2xy + y^2$ that the cross term $2p$ sits inside $s^2$ and must be subtracted. 3

2. A radical equation and the candidates it produces 9 points

Part A 3

Squares both sides, expanding the binomial in full, and forms the quadratic. 2
Lists both roots as candidates, not yet confirmed solutions. 1

Part B 3

Substitutes each candidate into the original equation, not the squared one. 2
Keeps only the candidate that checks, discarding the extraneous one. 1

Part C 3

Traces the extra candidate to the second branch squaring admits, and names the nonnegativity condition $x - 5 \geq 0$ the original imposes. 3

3. A fraction that contains a copy of itself 9 points

Part A 3

Identifies the tail below the first $6 +$ as the copy $\tfrac{1}{x}$ and writes $x = 6 + \tfrac{1}{x}$ . 2
Clears the fraction to a quadratic in $x$ . 1

Part B 3

Solves the quadratic by the formula or by completing the square. 2
Selects the positive root as the value, rejecting the negative one. 1

Part C 3

Reads $x = 6 + \tfrac{1}{x}$ as saying the number exceeds its reciprocal by exactly $6$ , tying the gap to the fraction's leading term. 3

4. Undoing even powers and the sign they hide 9 points

Part A 3

Takes the root of both sides and keeps the $\pm$ from the even root. 2
Reports both solutions, not only the positive branch. 1

Part B 3

Sets $u = x^{1/3}$ , finds $u = \pm 3$ , and cubes to recover $x$ . 2
Reports both values, the even numerator producing two. 1

Part C 3

Explains that an even power identifies $x$ with $-x$ , so both signs survive undoing it, while an odd power keeps them distinct. 3

5. A tower of square roots, named and checked 9 points

Part A 3

Replaces the inner copy by $x$ and writes $x = \sqrt{56 + x}$ . 2
Squares to reach a quadratic in $x$ . 1

Part B 3

Solves the quadratic to both roots. 2
Selects the nonnegative root as the value and discards the negative one. 1

Part C 3

Attributes the extra root to squaring admitting the negative branch, and rejects it by the nonnegativity of the original tower, as with an extraneous root. 3

6. When an infinite sum has a value and when it has none 9 points

Part A 3

Finds the common ratio of each series by dividing consecutive terms. 2
Applies $\lvert r \rvert < 1$ to decide which series settle. 1

Part B 3

Applies $\tfrac{a_1}{1 - r}$ with the sign of $r$ handled for each settling series. 2
Leaves the non-settling series unsummed rather than forcing a value. 1

Part C 3

Ties the formula to the assumption $S = a_1 + rS$ and explains that only $\lvert r \rvert < 1$ makes the partial sums settle, so only then is the output a true sum. 3

7. A system that hides its product 9 points

Part A 3

Rearranges $x^2 + y^2 = s^2 - 2p$ to solve for $p$ . 2
Identifies the result as the product $xy$ . 1

Part B 3

Forms $t^2 - st + p = 0$ from the sum and product and solves it. 2
Reports the result as the unordered pair. 1

Part C 3

Explains that the swap-invariance of the system yields only the unordered pair, and reads $s^2 - 4p > 0$ as the guarantee of a real, distinct pair. 3

8. Two manipulations that change the solution set 9 points

Part A 3

Solves $x^2 = 9$ to $x = \pm 3$ , keeping both signs. 2
Discards $x = 3$ as outside the original's domain, keeping only $x = -3$ . 1

Part B 3

Factors $x(x - 6) = 0$ instead of dividing by $x$ . 2
Names $x = 0$ as the solution that dividing by $x$ would lose. 1

Part C 3

Explains that a vanishing factor forces equality at its zero (adding a solution) while dividing by it drops the zero case (losing a solution). 3

9. A self-similar computation to audit 9 points

Part A 3

Locates the fault in the premise that the series has a sum, rather than in any single algebra step. 1
Shows the partial sums grow without bound, so no number $S$ exists to name. 2

Part B 3

Applies the self-similar move and notes $\lvert r \rvert < 1$ makes it valid. 2
Reports $T = \tfrac{3}{2}$ as a genuine value, not an empty one. 1

Part C 3

States the $\lvert r \rvert < 1$ condition and ties the trustworthiness of the identical algebra to whether the partial sums settle. 3

10. An equation whose coefficients read the same both ways 11 points

Part A 3

Checks that $x = 0$ is not a root, then divides by $x^2$ and groups the mirror terms. 2
Substitutes $x^2 + \tfrac{1}{x^2} = u^2 - 2$ to reach a quadratic in $u$ . 1

Part B 5

Solves the quadratic in $u$ for both values. 2
Back-substitutes each $u$ , solving $x + \tfrac{1}{x} = u$ for $x$ . 2
Reports all four values of $x$ . 1

Part C 3

Uses the palindrome identity $P(x) = x^n P(\tfrac{1}{x})$ to show $P(r) = 0$ forces $P(\tfrac{1}{r}) = 0$ , given $r \neq 0$ . 3

Multiple choice

Free response