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Radicals and Rational Exponents: 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, 77 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 routes to one exponent, and the arithmetic each one costs . 9 points. Question 1 of 10.

This question is about evaluating rational exponents by taking the root first and the power second, and about why that order never changes the answer.
1. Part A.
  Evaluate $216^{2/3}$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Evaluate $\left(\dfrac{1}{49}\right)^{-3/2}$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  The alternate route for part A computes $\sqrt[3]{216^2}$ instead of $\left(\sqrt[3]{216}\right)^2$ . Compute $216^2$ , and explain why this route reaches the same value as part A despite the extra arithmetic.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

$36$ .

Part B

$343$ .

Part C

$216^2 = 46656$ , and $\sqrt[3]{46656} = 36$ since $36^3 = 46656$ . Both routes agree because a positive base has exactly one positive cube root, so root-then-power and power-then-root always land on the same number; taking the root first only keeps the numbers small.

Worked solution

Part A

Take the cube root first, since the denominator of the exponent names the root, then square it.

$216^{2/3} = \left(\sqrt[3]{216}\right)^2 = 6^2 = 36$

Part B

A negative exponent means take the reciprocal first; only then apply the rational exponent.

$\left(\frac{1}{49}\right)^{-3/2} = 49^{3/2} = \left(\sqrt{49}\right)^3 = 7^3 = 343$

Part C

Squaring first gives $216^2 = 46656$ , and since $36^3 = 46656$ , the cube root of that is $36$ , matching part A.

$\sqrt[3]{216^2} = \sqrt[3]{46656} = 36$

The two routes must agree because $216$ is positive: a positive number has exactly one positive cube root, so $\left(\sqrt[3]{216}\right)^2$ and $\sqrt[3]{216^2}$ are two names for the same number, one of them just easier to compute by hand.

In one line

$216^{2/3}=36$ and $\left(\frac1{49}\right)^{-3/2}=343$ . Since $216^2=46656=36^3$ , the two routes for part A agree, because a positive base has exactly one positive cube root; taking the root first only keeps the arithmetic small, it never changes the result.

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 cube root before applying the outer power. . Worth 2 points.

Reports a single whole-number result with no remaining radical. . Worth 1 point.

Part B 3 points

Reciprocates the base to clear the negative exponent before applying the power. . Worth 2 points.

Reports a positive result, since a negative exponent flips a value over rather than flipping its sign. . Worth 1 point.

Part C 3 points

Grounds the agreement between the two routes in a positive base having exactly one positive $n$ -th root, not in coincidence. . Worth 2 points. needs an explanation, not just an answer

Computes $216^2 = 46656$ correctly and confirms $36^3 = 46656$ . . Worth 1 point.
2. One radicand that needs a restriction, and one that never does . 9 points. Question 2 of 10.

Both parts below ask for simplest radical form. Watch what each radicand's exponent on $x$ forces you to say about the domain.
1. Part A.
  Put $\sqrt{45x^3}$ in simplest radical form, given that $x \ge 0$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Put $\sqrt{75x^4y}$ in simplest radical form, where $x$ may be any real number and $y \ge 0$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  Part A required $x \ge 0$ before you could even write $\sqrt{45x^3}$ , while part B allowed $x$ to be any real number. Explain what feature of each radicand causes that difference.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

$3x\sqrt{5x}$ .

Part B

$5x^2\sqrt{3y}$ .

Part C

Part A's radicand carries $x^3$ , an ODD power; since the index is even, an odd power of a possibly negative $x$ could be negative, so nonnegativity had to be imposed. Part B's radicand carries $x^4$ , an EVEN power, automatically nonnegative for every real $x$ , so no restriction was needed.

Worked solution

Part A

Split off the largest perfect square factor: $45x^3 = 9x^2\cdot5x$ . Since $x\ge0$ , $\sqrt{x^2}=x$ with no bars needed.

$\sqrt{45x^3} = \sqrt{9x^2}\sqrt{5x} = 3x\sqrt{5x}$

Part B

Split off the perfect square factor $25x^4$ . Since the exponent emerging from the root is $2$ , an even number, no bars are needed regardless of the sign of $x$ .

$\sqrt{75x^4y} = \sqrt{25x^4}\sqrt{3y} = 5x^2\sqrt{3y}$

Part C

The index in both cases is $2$ (even), which forces the radicand to be nonnegative. In part A, the radicand's $x^3$ is an odd power of $x$ , so it takes the sign of $x$ ; a negative $x$ would make $45x^3$ negative, so $x \ge 0$ had to be given. In part B, the radicand's $x^4$ is an even power, which is $\ge 0$ for every real $x$ automatically, so the radicand $75x^4y$ never needs $x$ restricted (only $y \ge 0$ was needed).

$45x^3 = 45x^2\cdot x, \text{ sign of } x; \qquad 75x^4y = 75x^4\cdot y, \ x^4 \ge 0 \text{ always}$

In one line

$\sqrt{45x^3}=3x\sqrt{5x}$ for $x\ge0$ , and $\sqrt{75x^4y}=5x^2\sqrt{3y}$ for any real $x$ (with $y\ge0$ ). Part A needed the restriction because $x^3$ is an odd power that could be negative; part B did not, because $x^4$ is always nonnegative.

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

Extracts the perfect square factor $9x^2$ , leaving $5x$ inside the radical. . Worth 2 points.

Uses the given $x \ge 0$ to write $\sqrt{x^2}=x$ without bars. . Worth 1 point.

Part B 3 points

Extracts the perfect square factor $25x^4$ , leaving $3y$ inside the radical. . Worth 2 points.

Reports $x^2$ with no absolute value bars, since the exponent emerging from the root is even. . Worth 1 point.

Part C 3 points

Identifies the odd exponent on $x$ in part A versus the even exponent in part B as the deciding feature. . Worth 2 points. needs an explanation, not just an answer

Ties the observation back to the even index needing a nonnegative radicand. . Worth 1 point.
3. A shipping bill that reads like a radical equation . 7 points. Question 3 of 10.

A shipping company charges $C = 15 + \sqrt{4x+9}$ dollars to ship a package weighing $x$ pounds.
1. Part A.
  A customer is billed $24. Isolate the radical, raise both sides to the matching power, and find $x$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  A different customer says they were billed $12 using the same pricing rule. Determine whether any weight $x$ could produce that bill, and justify your answer without doing any unnecessary algebra.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$x=18$ pounds.

Part B

No such weight exists: solving would require $\sqrt{4x+9}=12-15=-3$ , and a principal square root can never equal a negative number, for any $x$ .

Worked solution

Part A

Isolate the radical, then square both sides.

$15+\sqrt{4x+9}=24 \;\Longrightarrow\; \sqrt{4x+9}=9 \;\Longrightarrow\; 4x+9=81 \;\Longrightarrow\; x=18$

Check: $4(18)+9=81$ , $\sqrt{81}=9$ , and $15+9=24$ , matching the bill.

Part B

Isolate the radical the same way as in part A. The left side is a principal square root, which is never negative, while the right side is $-3$ . No value of $x$ can make a nonnegative quantity equal a negative one, so the equation has no solution, and no further algebra (squaring, solving) is needed to see it.

$15+\sqrt{4x+9}=12 \;\Longrightarrow\; \sqrt{4x+9} = -3 \quad (\text{impossible, since } \sqrt{4x+9}\ge0)$

In one line

The $24 bill corresponds to $x=18$ pounds. A $12 bill is impossible: isolating the radical would require $\sqrt{4x+9}=-3$ , and a principal square root is never negative, so no weight produces that bill.

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

Isolates the radical before squaring. . Worth 1 point.

Squares and solves the resulting linear equation correctly. . Worth 2 points.

Reports the solved weight with the unit pounds. . Worth 1 point.

Part B 3 points

Isolates the radical to see that it would need to equal $-3$ . . Worth 2 points. needs an explanation, not just an answer

States plainly that a principal square root is never negative, so no further algebra is needed. . Worth 1 point.
4. The same two radicals, multiplied two different ways . 9 points. Question 4 of 10.

Both parts multiply expressions built from $\sqrt2$ and $\sqrt5$ . Only one of the two products is a true conjugate pair.
1. Part A.
  Expand $(3\sqrt2 + \sqrt5)(2\sqrt2 - \sqrt5)$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Expand $(3\sqrt2 + \sqrt5)(3\sqrt2 - \sqrt5)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Explain why part B's product came out rational while part A's did not, referring to the coefficients in each expression.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

$7 - \sqrt{10}$ .

Part B

$13$ .

Part C

In part B the two binomials are true conjugates, both carry $3\sqrt2$ , so the cross terms $-3\sqrt2\sqrt5$ and $+3\sqrt2\sqrt5$ are exact opposites and cancel. In part A the second factor uses $2\sqrt2$ , not $3\sqrt2$ , so the cross terms are unequal in size and only partly cancel, leaving $-\sqrt{10}$ behind.

Worked solution

Part A

Expand every term against every term, then use $\sqrt2\cdot\sqrt2=2$ and $\sqrt5\cdot\sqrt5=5$ .

$(3\sqrt2+\sqrt5)(2\sqrt2-\sqrt5) = 6(2) - 3\sqrt{10} + 2\sqrt{10} - 5 = (12-5) + (-3+2)\sqrt{10} = 7-\sqrt{10}$

Part B

This is a true conjugate pair: the matching term is $3\sqrt2$ in both factors, so the cross terms are exact opposites and cancel.

$(3\sqrt2+\sqrt5)(3\sqrt2-\sqrt5) = (3\sqrt2)^2 - (\sqrt5)^2 = 18 - 5 = 13$

Part C

A conjugate pair needs the SAME coefficient on the matching term in both factors, so that the two cross terms are exact opposites. In part B both factors carry $3\sqrt2$ , so the cross terms cancel completely; in part A the second factor carries $2\sqrt2$ instead of $3\sqrt2$ , so they do not.

$\text{Part B: } -3\sqrt2\sqrt5 + 3\sqrt2\sqrt5 = 0, \qquad \text{Part A: } -3\sqrt{10} + 2\sqrt{10} = -\sqrt{10}$

In one line

$(3\sqrt2+\sqrt5)(2\sqrt2-\sqrt5)=7-\sqrt{10}$ , while the true conjugate pair $(3\sqrt2+\sqrt5)(3\sqrt2-\sqrt5)=13$ . Only matching coefficients on the shared term make the cross terms exact opposites, which is what cancels the radical.

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

Expands all four products and resolves each radical-times-radical term. . Worth 2 points.

Collects the two cross terms into a single radical term rather than leaving them separate. . Worth 1 point.

Part B 3 points

Recognizes the conjugate pattern and applies the difference of squares directly. . Worth 1 point.

Squares both terms in the conjugate pair, including the coefficient on the radical term. . Worth 1 point.

Reports the result as a single rational number, with no radical remaining. . Worth 1 point.

Part C 3 points

Identifies that a true conjugate pair needs identical coefficients on the matching term for the cross terms to be exact opposites. . Worth 2 points. needs an explanation, not just an answer

Cites the specific cross terms from each part to support the explanation. . Worth 1 point.
5. Why an even power can never reach a negative target . 6 points. Question 5 of 10.

This question builds, from scratch, the argument behind the chapter's central restriction: why an even-indexed root of a negative number is never real.
1. Part A.
  Let $n$ be an even positive integer and let $r$ be any real number. Show that $r^n \ge 0$ .
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 3 points
2. Part B.
  Use part A to prove that if $a < 0$ and $n$ is even, no real number $r$ satisfies $r^n = a$ .
  
  Carry your own answer forward Use whatever form you reached in part A, even if it was not exactly $r^n=(r^m)^2$ ; the credit here is for combining a nonnegative-square fact with $a<0$ to reach a contradiction.
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 3 points
The answer

Part A

For even $n=2m$ , $r^n=(r^m)^2$ , and the square of any real number is $\ge0$ . So $r^n\ge0$ for every real $r$ whenever $n$ is even.

Part B

If $a<0$ , no real $r$ satisfies $r^n=a$ : part A shows $r^n\ge0$ for every real $r$ when $n$ is even, but $a<0$ , so $r^n=a$ would force a nonnegative number to equal a negative one, which is impossible.

Worked solution

Part A

Since $n$ is even, write $n=2m$ for a positive integer $m$ . The square of any real number is nonnegative, regardless of the sign of $r^m$ , so $r^n \ge 0$ .

$r^n = r^{2m} = (r^m)^2 \ge 0$

Part B

Suppose, for contradiction, that some real $r$ satisfies $r^n=a$ with $n$ even and $a<0$ . By part A, $r^n \ge 0$ , so this would force $a \ge 0$ too, contradicting $a<0$ . So no such $r$ exists.

$r^n = a < 0 \quad \text{contradicts} \quad r^n \ge 0 \text{ (part A)}$

This is exactly why $\sqrt[n]{a}$ names no real number when $a<0$ and $n$ is even.

In one line

For even $n$ , write $n=2m$ ; then $r^n=(r^m)^2\ge0$ for every real $r$ . Since $a<0$ can never equal a nonnegative number, no real $r$ satisfies $r^n=a$ . This is exactly why $\sqrt[n]{a}$ names no real number when $a<0$ and $n$ is even.

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

Writes $n=2m$ to use the evenness of $n$ . . Worth 1 point.

Concludes $r^n=(r^m)^2$ is a square of a real number, hence nonnegative. . Worth 2 points. needs an explanation, not just an answer

Part B 3 points

Combines part A's $r^n \ge 0$ with $a<0$ to reach a contradiction. . Worth 2 points. needs an explanation, not just an answer

States the conclusion clearly, connecting it back to why $\sqrt[n]{a}$ is undefined for $a<0$ and even $n$ . . Worth 1 point.
6. One ramp slope, rationalized two ways . 6 points. Question 6 of 10.

A ramp's exact slope, in feet of rise per foot of run, is $\dfrac{9}{\sqrt{20}}$ . A contractor needs it written with a whole-number denominator.
1. Part A.
  Rationalize the denominator of $\dfrac{9}{\sqrt{20}}$ , giving the answer in lowest terms.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  A coworker instead multiplies top and bottom by $\sqrt5$ alone, since $\sqrt{20}=2\sqrt5$ . Show that this shortcut reaches the same answer, and explain why it is legitimate even though $\sqrt5$ is not the original denominator.
  
  Carry your own answer forward Compare the shortcut's result with whatever value you found in part A; use that result consistently in your comparison.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 3 points
The answer

Part A

$\dfrac{9\sqrt5}{10}$ .

Part B

Multiplying by $\sqrt5/\sqrt5$ gives $\dfrac{9\sqrt5}{2\cdot5}=\dfrac{9\sqrt5}{10}$ , the same result, since $\sqrt5$ , not $\sqrt{20}$ , is the factor that completes the square once $\sqrt{20}$ is simplified to $2\sqrt5$ .

Worked solution

Part A

Multiply top and bottom by $\sqrt{20}$ , then simplify.

$\frac{9}{\sqrt{20}}\cdot\frac{\sqrt{20}}{\sqrt{20}} = \frac{9\sqrt{20}}{20} = \frac{9\cdot2\sqrt5}{20} = \frac{18\sqrt5}{20} = \frac{9\sqrt5}{10}$

Part B

Simplifying first, $\frac{9}{\sqrt{20}} = \frac{9}{2\sqrt5}$ . It is legitimate because once $\sqrt{20}$ is written as $2\sqrt5$ , the factor $\sqrt5$ , not $\sqrt{20}$ , is what genuinely completes the square in the denominator; multiplying by either form of $1$ clears the radical.

$\frac{9}{2\sqrt5}\cdot\frac{\sqrt5}{\sqrt5} = \frac{9\sqrt5}{2(5)} = \frac{9\sqrt5}{10}$

In one line

$\dfrac{9}{\sqrt{20}} = \dfrac{9\sqrt5}{10}$ . Multiplying instead by $\sqrt5/\sqrt5$ , after first simplifying $\sqrt{20}=2\sqrt5$ , reaches the identical fraction, because $\sqrt5$ is the factor that actually completes the square once the radical itself is simplified.

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

Multiplies top and bottom by $\sqrt{20}$ (or an equivalent completing factor). . Worth 2 points.

Reduces the resulting fraction to lowest terms. . Worth 1 point.

Part B 3 points

Shows the shortcut reaches the identical fraction found in part A. . Worth 2 points.

Explains why multiplying by the simplified radical still legitimately rationalizes the denominator. . Worth 1 point. needs an explanation, not just an answer
7. A shortcut that squares a difference the wrong way . 7 points. Question 7 of 10.

This question is about $\sqrt{2x+6} - \sqrt{x+1} = 1$ , solved correctly and then compared with a classmate's flawed shortcut.
1. Part A.
  Solve $\sqrt{2x+6} - \sqrt{x+1} = 1$ , isolating one radical at a time and raising to the matching power at each step.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  A classmate instead squares both sides directly, writing $(2x+6)-(x+1)=1^2$ , and concludes $x=-4$ . Identify the specific algebra error, and explain why $x=-4$ could never have been a genuine solution regardless of the error.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 3 points
The answer

Part A

no real solution (the quadratic $x^2+4x+12=0$ has a negative discriminant).

Part B

The student squared term by term, using $(A-B)^2=A^2-B^2$ , which is false; squaring a difference must expand the cross term. Also $x=-4$ fails the domain, since $x+1=-3<0$ makes $\sqrt{x+1}$ not even real.

Worked solution

Part A

Isolate one radical, square, isolate the surviving radical, and square again.

$\sqrt{2x+6}=1+\sqrt{x+1} \;\Longrightarrow\; x+4=2\sqrt{x+1} \;\Longrightarrow\; (x+4)^2=4(x+1) \;\Longrightarrow\; x^2+4x+12=0$

The discriminant is $16-48=-32$ , which is negative, so there is no real candidate at all.

Part B

The classmate's step treats $(\sqrt{2x+6}-\sqrt{x+1})^2$ as $(2x+6)-(x+1)$ , which is the false rule $(A-B)^2=A^2-B^2$ ; the correct expansion keeps a cross term.

$(A-B)^2 = A^2 - 2AB + B^2 \ne A^2 - B^2$

Independent of that error, $x=-4$ gives $x+1=-3<0$ , so $\sqrt{x+1}$ is not even a real number there, meaning $x=-4$ could never have satisfied the original equation.

In one line

$\sqrt{2x+6}-\sqrt{x+1}=1$ has no real solution: isolating and squaring twice leads to $x^2+4x+12=0$ , whose discriminant is negative. A classmate's shortcut of squaring term by term, $(2x+6)-(x+1)=1$ , is not a legal expansion of $(A-B)^2$ , and its candidate $x=-4$ also falls outside the domain.

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

Isolates a single radical before each squaring step. . Worth 1 point.

Reaches $x^2+4x+12=0$ and computes its discriminant correctly. . Worth 2 points.

Recognizes that a negative discriminant means the equation has no real solution at all. . Worth 1 point.

Part B 3 points

Names the term-by-term squaring error specifically, contrasting it with the correct expansion of $(A-B)^2$ . . Worth 2 points. needs an explanation, not just an answer

Notes that $x=-4$ falls outside the domain independent of the squaring error. . Worth 1 point.
8. A formula that stops being real before the algebra says so . 9 points. Question 8 of 10.

A physics model gives the speed (in meters per second) of a wave on a string as $v = \sqrt{45 - 5x^2}$ , where $x$ is a tension parameter.
1. Part A.
  For what values of $x$ does $v=\sqrt{45-5x^2}$ represent a real speed?
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Evaluate $v$ at $x=1$ , in simplest radical form.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Suppose someone evaluates $v$ at $x=4$ without first checking the domain from part A. Explain what goes wrong, referencing the parity of the index.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

$-3 \le x \le 3$ .

Part B

$2\sqrt{10}$ m/s.

Part C

At $x=4$ the radicand is $45-5(16)=-35$ , which is negative. Since the index is even, a negative radicand names no real number, so $v$ is undefined there, exactly why part A restricted $x$ to $[-3,3]$ .

Worked solution

Part A

An even-indexed root needs a nonnegative radicand.

$45-5x^2 \ge 0 \;\Longrightarrow\; x^2 \le 9 \;\Longrightarrow\; -3 \le x \le 3$

Part B

Substitute and extract the perfect square factor from the radicand.

$v = \sqrt{45-5(1)^2} = \sqrt{40} = \sqrt{4\cdot10} = 2\sqrt{10} \text{ m/s}$

Part C

Because $v$ is defined by a square root, an even-indexed root, a negative radicand means the expression names no real number: $v$ is simply undefined at $x=4$ . This is exactly the boundary that part A's domain restriction, $-3 \le x \le 3$ , was set up to exclude.

$45 - 5(4)^2 = 45 - 80 = -35 < 0 \quad \Longrightarrow \quad v = \sqrt{-35} \text{ is not real}$

In one line

The formula is real for $-3\le x\le3$ ; at $x=1$ , $v=\sqrt{40}=2\sqrt{10}$ m/s. At $x=4$ the radicand is $45-80=-35<0$ , so the even-indexed root gives no real number there, which is exactly the boundary part A found.

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

Sets up the inequality $45-5x^2 \ge 0$ . . Worth 1 point.

Solves the domain inequality and reports the resulting interval. . Worth 2 points.

Part B 3 points

Substitutes the given parameter and simplifies the radicand correctly. . Worth 1 point.

Extracts the greatest perfect-square factor, simplifies fully, and reports the result with the correct units. . Worth 2 points.

Part C 3 points

Computes the radicand at $x=4$ and identifies it as negative. . Worth 1 point. needs an explanation, not just an answer

Connects the even index to why a negative radicand fails to be real, tying it back to part A's restriction. . Worth 2 points. needs an explanation, not just an answer
9. Why a cube root needs more than a sign flip . 6 points. Question 9 of 10.

A student wants to rationalize $\dfrac{1}{\sqrt[3]7-2}$ and reaches for the ordinary conjugate, multiplying by $\sqrt[3]7+2$ .
1. Part A.
  Compute $(\sqrt[3]7-2)(\sqrt[3]7+2)$ , and state whether the denominator is now free of radicals.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Explain, in general terms using $x=\sqrt[3]d$ for an irrational cube root (where $d$ is rational but not a perfect cube) and $a$ for a rational number, why a two-term sign flip can never fully clear such a cube root, while it always clears a square root.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$\sqrt[3]{49}-4$ , which still contains a radical, so the denominator is not yet rational.

Part B

A square root's square, $d$ , is rational, so $(a+\sqrt d)(a-\sqrt d)=a^2-d$ needs only one cancellation. A cube root's square, $x^2=\sqrt[3]{d^2}$ , is ALSO irrational, so $(a+x)(a-x)=a^2-x^2$ leaves that term behind; only $x^3$ is rational.

Worked solution

Part A

Expand as a difference of squares.

$(\sqrt[3]7-2)(\sqrt[3]7+2) = (\sqrt[3]7)^2 - 4 = \sqrt[3]{49} - 4$

Since $49$ is not a perfect cube, $\sqrt[3]{49}$ is irrational, so the denominator is still not rational.

Part B

For a square root $\sqrt d$ , the relation $(\sqrt d)^2=d$ is rational, so one sign flip already clears it. For a cube root $x=\sqrt[3]d$ , the square $x^2=\sqrt[3]{d^2}$ is ALSO irrational (only $x^3=d$ is rational), so a sign flip leaves that term behind.

$(a+\sqrt d)(a-\sqrt d) = a^2 - d \ \text{(rational)}, \qquad (a+x)(a-x) = a^2 - x^2 \ \text{(still irrational)}$

A sign flip only ever cancels an ODD power of the new object; a cube root has two irrational powers standing in the way, not one.

In one line

$(\sqrt[3]7-2)(\sqrt[3]7+2)=\sqrt[3]{49}-4$ , still irrational, so the ordinary conjugate fails. In general, a square root's square is rational, but a cube root's square is still irrational, so a two-term sign flip leaves that squared term behind; only the cube itself is rational, which is why a three-term multiplier is needed.

Rubric

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

Part A 3 points

Expands the product using the difference-of-squares pattern. . Worth 2 points.

Determines whether a radical remains and states whether the resulting denominator is rational. . Worth 1 point.

Part B 3 points

Identifies that an irrational cube root's square is also irrational, unlike a square root's square. . Worth 2 points. needs an explanation, not just an answer

States the general pattern clearly in terms of $a$ and $x$ . . Worth 1 point.
10. Welding two pipe lengths, and the rule that would have gone wrong . 9 points. Question 10 of 10.

A fabricator is joining two lengths of pipe cut from the same stock: side lengths $\sqrt{18}$ and $\sqrt8$ meters for one frame, and widths $\sqrt{50}$ and $\sqrt{98}$ meters for another.
1. Part A.
  The frame's area is the product of its two side lengths, $\sqrt{18}\cdot\sqrt8$ . Evaluate it.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  The two widths, $\sqrt{50}$ and $\sqrt{98}$ meters, need to be welded end to end. Find the total length in simplest form.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Part A's product rule needs $a\ge0$ and $b\ge0$ to guarantee it lands on the correct nonnegative root. Explain why $\sqrt{-18}\cdot\sqrt{-8}$ could NOT be evaluated by the same product-rule step used in part A, and state what it equals instead.
  
  Carry your own answer forward Compare your complex-number result with whatever value you found in part A; use your part A result consistently in the comparison.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 3 points
The answer

Part A

$12$ m $^2$ .

Part B

$12\sqrt2$ meters.

Part C

Both radicands are negative, so the product rule's hypothesis fails. Converting to $i$ -form first, $(i\sqrt{18})(i\sqrt8)=i^2\sqrt{144}=-12$ (m $^2$ in the analogous units), the NEGATIVE of part A's value, not $12$ .

Worked solution

Part A

Both radicands are nonnegative, so the product rule applies directly.

$\sqrt{18}\cdot\sqrt8 = \sqrt{144} = 12 \text{ m}^2$

Part B

Simplify each radical to expose a shared unit, then add the like radicals.

$\sqrt{50}+\sqrt{98} = 5\sqrt2 + 7\sqrt2 = 12\sqrt2 \text{ meters}$

Part C

The product rule $\sqrt a\sqrt b=\sqrt{ab}$ requires $a\ge0$ and $b\ge0$ ; here both radicands are negative, so the hypothesis fails and the rule cannot be applied directly. Converting each factor to $i$ -form first,

$\sqrt{-18}\cdot\sqrt{-8} = (i\sqrt{18})(i\sqrt8) = i^2\sqrt{144} = -12,$

which is the negative of part A's answer, not the $12$ that blindly applying the product rule to $\sqrt{(-18)(-8)}=\sqrt{144}$ would give.

In one line

$\sqrt{18}\cdot\sqrt8=12$ m $^2$ and $\sqrt{50}+\sqrt{98}=12\sqrt2$ meters. But $\sqrt{-18}\cdot\sqrt{-8}$ cannot use the same product-rule step, since both radicands are negative; converting to $i$ -form first gives $(i\sqrt{18})(i\sqrt8)=-12$ , the negative of part A's 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

Applies the product rule to combine the two radicands into a single radical. . Worth 2 points.

Simplifies the radical completely and reports the result with the correct units. . Worth 1 point.

Part B 3 points

Simplifies each radical to extract its perfect square factor. . Worth 2 points.

Recognizes the two radicals share the radicand $\sqrt2$ and combines them into a single term. . Worth 1 point.

Part C 3 points

Identifies that both radicands are negative, so the product rule's hypothesis fails. . Worth 2 points. needs an explanation, not just an answer

Converts both radicals to $i$ -form, accounts for the product of the imaginary units, and simplifies completely. . 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 routes to one exponent, and the arithmetic each one costs 9 points

Part A 3

Takes the cube root before applying the outer power. 2
Reports a single whole-number result with no remaining radical. 1

Part B 3

Reciprocates the base to clear the negative exponent before applying the power. 2
Reports a positive result, since a negative exponent flips a value over rather than flipping its sign. 1

Part C 3

Grounds the agreement between the two routes in a positive base having exactly one positive $n$ -th root, not in coincidence. 2
Computes $216^2 = 46656$ correctly and confirms $36^3 = 46656$ . 1

2. One radicand that needs a restriction, and one that never does 9 points

Part A 3

Extracts the perfect square factor $9x^2$ , leaving $5x$ inside the radical. 2
Uses the given $x \ge 0$ to write $\sqrt{x^2}=x$ without bars. 1

Part B 3

Extracts the perfect square factor $25x^4$ , leaving $3y$ inside the radical. 2
Reports $x^2$ with no absolute value bars, since the exponent emerging from the root is even. 1

Part C 3

Identifies the odd exponent on $x$ in part A versus the even exponent in part B as the deciding feature. 2
Ties the observation back to the even index needing a nonnegative radicand. 1

3. A shipping bill that reads like a radical equation 7 points

Part A 4

Isolates the radical before squaring. 1
Squares and solves the resulting linear equation correctly. 2
Reports the solved weight with the unit pounds. 1

Part B 3

Isolates the radical to see that it would need to equal $-3$ . 2
States plainly that a principal square root is never negative, so no further algebra is needed. 1

4. The same two radicals, multiplied two different ways 9 points

Part A 3

Expands all four products and resolves each radical-times-radical term. 2
Collects the two cross terms into a single radical term rather than leaving them separate. 1

Part B 3

Recognizes the conjugate pattern and applies the difference of squares directly. 1
Squares both terms in the conjugate pair, including the coefficient on the radical term. 1
Reports the result as a single rational number, with no radical remaining. 1

Part C 3

Identifies that a true conjugate pair needs identical coefficients on the matching term for the cross terms to be exact opposites. 2
Cites the specific cross terms from each part to support the explanation. 1

5. Why an even power can never reach a negative target 6 points

Part A 3

Writes $n=2m$ to use the evenness of $n$ . 1
Concludes $r^n=(r^m)^2$ is a square of a real number, hence nonnegative. 2

Part B 3

Combines part A's $r^n \ge 0$ with $a<0$ to reach a contradiction. 2
States the conclusion clearly, connecting it back to why $\sqrt[n]{a}$ is undefined for $a<0$ and even $n$ . 1

6. One ramp slope, rationalized two ways 6 points

Part A 3

Multiplies top and bottom by $\sqrt{20}$ (or an equivalent completing factor). 2
Reduces the resulting fraction to lowest terms. 1

Part B 3

Shows the shortcut reaches the identical fraction found in part A. 2
Explains why multiplying by the simplified radical still legitimately rationalizes the denominator. 1

7. A shortcut that squares a difference the wrong way 7 points

Part A 4

Isolates a single radical before each squaring step. 1
Reaches $x^2+4x+12=0$ and computes its discriminant correctly. 2
Recognizes that a negative discriminant means the equation has no real solution at all. 1

Part B 3

Names the term-by-term squaring error specifically, contrasting it with the correct expansion of $(A-B)^2$ . 2
Notes that $x=-4$ falls outside the domain independent of the squaring error. 1

8. A formula that stops being real before the algebra says so 9 points

Part A 3

Sets up the inequality $45-5x^2 \ge 0$ . 1
Solves the domain inequality and reports the resulting interval. 2

Part B 3

Substitutes the given parameter and simplifies the radicand correctly. 1
Extracts the greatest perfect-square factor, simplifies fully, and reports the result with the correct units. 2

Part C 3

Computes the radicand at $x=4$ and identifies it as negative. 1
Connects the even index to why a negative radicand fails to be real, tying it back to part A's restriction. 2

9. Why a cube root needs more than a sign flip 6 points

Part A 3

Expands the product using the difference-of-squares pattern. 2
Determines whether a radical remains and states whether the resulting denominator is rational. 1

Part B 3

Identifies that an irrational cube root's square is also irrational, unlike a square root's square. 2
States the general pattern clearly in terms of $a$ and $x$ . 1

10. Welding two pipe lengths, and the rule that would have gone wrong 9 points

Part A 3

Applies the product rule to combine the two radicands into a single radical. 2
Simplifies the radical completely and reports the result with the correct units. 1

Part B 3

Simplifies each radical to extract its perfect square factor. 2
Recognizes the two radicals share the radicand $\sqrt2$ and combines them into a single term. 1

Part C 3

Identifies that both radicands are negative, so the product rule's hypothesis fails. 2
Converts both radicals to $i$ -form, accounts for the product of the imaginary units, and simplifies completely. 1

Multiple choice

Free response