Algebra 3 Quiz: Radical Expressions

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Radical ExpressionsQuestion 1 of 20

Simplify $\left(\frac{27x^9}{y^{-6}}\right)^{2/3}$ . Assume $x,y\neq0$ .

9x^6 y^4

9x^6 y^{-4}

3x^6 y^4

9x^6 y^2

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Algebra 3 Quiz

Algebra 3 Quiz: Radical Expressions

Practice Radical Expressions in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Radical Expressions, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Simplify $\left(\frac{27x^9}{y^{-6}}\right)^{2/3}$ . Assume $x,y\neq0$ .

$9x^6 y^4$ (correct answer)
$9x^6 y^{-4}$
$3x^6 y^4$
$9x^6 y^2$

Explanation: When you see a rational expression raised to a power, simplify inside first using exponent rules—especially negative exponents. Here,

y^{-6}

in the denominator means the expression is equivalent to

27x^9 y^6

. Now apply the outer exponent

\frac{2}{3}

to each factor:

27^{2/3} = (3^3)^{2/3} = 3^2 = 9

x^{9 \cdot \frac{2}{3}} = x^6

, and

y^{6 \cdot \frac{2}{3}} = y^4

. Multiplying gives

9x^6 y^4

, which is the correct simplification. The choice

9x^6 y^{-4}

shows the classic trap of applying the exponent to

y^{-6}

as if it were in the numerator:

(y^{-6})^{2/3} = y^{-4}

, but that ignores the denominator—negative exponents already indicate a reciprocal, so

y^{-6}

moves up as

y^6

before raising to

\frac{2}{3}

. The choice

3x^6 y^4

comes from computing

27^{2/3}

3

instead of

9

—you must take the cube root first, then square. The choice

9x^6 y^2

treats the exponent on

y

6 \times \frac{2}{3} = 4

, so

y^2

indicates a miscalculation or confusion of exponent rules. Your study tip: for any expression

(a^m b^n)^p

, distribute

p

to each exponent:

a^{mp}b^{np}

. And remember that a negative exponent in the denominator means the base belongs in the numerator before applying any outside power. This two-step habit prevents the most common traps.

Question 2

For all real $x$ , which expression is equivalent to $\sqrt{25x^2-10x+1}$ ?

$5x-1$
$|5x-1|$ (correct answer)
$1-5x$
$\sqrt{5x-1}$

Explanation: When you see a square root containing a quadratic like

25x^2-10x+1

, your first instinct should be to check whether the quadratic is a perfect square trinomial. Here,

25x^2-10x+1 = (5x-1)^2

, so the expression becomes

\sqrt{(5x-1)^2}

. Now the key point: for any real number,

\sqrt{a^2} = |a|

, not

a

. That is because the principal square root is always nonnegative, while

5x-1

can be negative. Therefore the equivalent expression is

|5x-1|

. Why are the other choices traps?

5x-1

ignores the absolute value; it fails when

x<0.2

, since then

5x-1

is negative and cannot equal a square root.

1-5x

is the negative of

5x-1

, so it only matches the square root when

5x-1

is negative; when

5x-1

is positive, it is wrong.

\sqrt{5x-1}

is not equivalent because it has a restricted domain and equals something entirely different; for example, at

x=0

, the original equals

1

but

\sqrt{5x-1}

is not even real. A quick check with a number can save you: plug in

x=0

. The original gives

\sqrt{1}=1

, while

5x-1=-1

and

1-5x=1

, so only the absolute-value form works for all

x

. Remember the rule: whenever you take the square root of something squared, write absolute value bars. That one habit prevents a whole class of algebra errors.

Question 3

Simplify $\sqrt{2}(\sqrt{8}-\sqrt{6})$ .

$4-2\sqrt{6}$
$4-\sqrt{3}$
$4-2\sqrt{3}$ (correct answer)
$4-6\sqrt{2}$

Explanation: When you see a radical multiplied by a binomial in parentheses, your first move should be to distribute the radical across the terms. This question tests whether you can multiply square roots using

\sqrt{a}\sqrt{b}=\sqrt{ab}

and then simplify the result by pulling out perfect square factors. Distribute:

\sqrt{2}(\sqrt{8}-\sqrt{6}) = \sqrt{2}\sqrt{8} - \sqrt{2}\sqrt{6} = \sqrt{16} - \sqrt{12}.

Since

\sqrt{16}=4

and

\sqrt{12}=\sqrt{4\cdot 3}=2\sqrt{3}

, the expression simplifies to

4-2\sqrt{3}

. Now, why are the other choices traps? The choice

4-2\sqrt{6}

comes from simplifying

\sqrt{12}

incorrectly as

2\sqrt{6}

. But

\sqrt{12}

is not

2\sqrt{6}

because

(2\sqrt{6})^2=24

, not 12. The choice

4-\sqrt{3}

shows the right radical, but it drops the factor of 2 from

2\sqrt{3}

. The choice

4-6\sqrt{2}

likely comes from "simplifying"

\sqrt{12}

6\sqrt{2}

, which is a serious error: you can only pull out the square root of a perfect-square factor, and 12 factors as

4\cdot 3

, not

6\cdot 2

in that way. A great habit: after simplifying a radical, square your simplified form to check that it matches the original radicand. For example,

(2\sqrt{3})^2 = 4\cdot 3 = 12

, confirming the simplification is correct.

Question 4

Which expression is equivalent to $\frac{2\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ ?

$3-\sqrt{15}$
$\sqrt{15}-3$ (correct answer)
$2\sqrt{15}-6$
$\sqrt{15}+3$

Explanation: When you see a fraction with radicals in the denominator, your first move should be to rationalize it: multiply the numerator and denominator by the conjugate of the denominator. Here, the conjugate of

\sqrt{5}+\sqrt{3}

\sqrt{5}-\sqrt{3}

. Doing this gives

\frac{2\sqrt{3}}{\sqrt{5}+\sqrt{3}}\cdot \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} = \frac{2\sqrt{3}(\sqrt{5}-\sqrt{3})}{(\sqrt{5})^2-(\sqrt{3})^2}.

Simplify the numerator:

2\sqrt{3}\sqrt{5}-2\sqrt{3}\sqrt{3}=2\sqrt{15}-6

. The denominator is the difference of squares:

5-3=2

. So you get

\frac{2\sqrt{15}-6}{2}=\sqrt{15}-3.

That is the expression you want. Now look at the traps:

2\sqrt{15}-6

is the numerator before you divided by the denominator's

2

, so it is not fully simplified.

3-\sqrt{15}

is the correct magnitude but with the wrong sign — a common error when subtracting in the numerator. And

\sqrt{15}+3

has the wrong sign on the

3

, usually from mishandling the conjugate subtraction. So the equivalent expression is

\sqrt{15}-3

. Study tip: after rationalizing, always check two things — did you simplify the fraction completely, and did the signs come out correctly? Writing the numerator and denominator separately before simplifying helps you avoid both traps.

Question 5

Simplify $\frac{6}{\sqrt[3]{9}}$ .

$2\sqrt[3]{3}$ (correct answer)
$6\sqrt[3]{3}$
$2\sqrt[3]{9}$
$\frac{2\sqrt[3]{3}}{3}$

Explanation: Whenever you see a cube root in the denominator, ask: what factor would make the radicand a perfect cube? Since

9=3^2

, multiplying by

\sqrt[3]{3}

completes the cube:

\sqrt[3]{9}\cdot\sqrt[3]{3}}=\sqrt[3]{27}=3.

So rationalize:

\frac{6}{\sqrt[3]{9}}}\cdot\frac{\sqrt[3]{3}}{\sqrt[3]{3}} = \frac{6\sqrt[3]{3}}}{3}=2\sqrt[3]{3}.

That is the simplified form. The choice

6\sqrt[3]{3}

comes from multiplying the numerator by

\sqrt[3]{3}

, but forgetting that the denominator becomes

3

, not

1

2\sqrt[3]{9}}

usually means someone confused

\sqrt[3]{9}

with

\sqrt{9}=3

, then wrote

2

with a leftover cube root; but

\sqrt[3]{9}\neq3

.And

\frac{2\sqrt[3]{3}}}{3}

is a trap after the rationalization step: since

\frac{6\sqrt[3]{3}}}{3}

, you should reduce

\frac{6}{3}=2

, giving

2\sqrt[3]{3}

. Do not leave the

3

in the denominator once you've divided the coefficient. Study tip: When rationalizing cube roots, multiply by the missing factor to create a perfect cube, then simplify the resulting fraction completely. A quick decimal check helps:

\sqrt[3]{9}\approx2.08

, so the original is about

2.88

, and

2\sqrt[3]{3}\approx2.88

Question 6

If $x=\frac{1}{\sqrt{3}-1}$ , what is the value of $x^2-2x$ ?

$\frac{\sqrt{3}}{2}$
$-\frac{\sqrt{3}}{2}$ (correct answer)
$\frac{1-\sqrt{3}}{2}$
$-\sqrt{3}$

Explanation: Whenever you see a variable defined with a radical in the denominator, rationalize before doing anything else. Multiply by the conjugate:

x=\frac{1}{\sqrt{3}-1}\cdot\frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{\sqrt{3}+1}{2}.

Now compute the needed pieces. Since

x^2=\frac{(\sqrt{3}+1)^2}{4}=\frac{4+2\sqrt{3}}{4}=\frac{2+\sqrt{3}}{2}

and

2x=\sqrt{3}+1=\frac{2\sqrt{3}+2}{2},

subtract:

x^2-2x=\frac{2+\sqrt{3}-(2\sqrt{3}+2)}{2}=-\frac{\sqrt{3}}{2}.

That is the correct value. The choice

\frac{\sqrt{3}}{2}

is the positive version of the answer; you would get it if you accidentally computed

2x-x^2

or dropped the negative sign during subtraction. The choice

\frac{1-\sqrt{3}}{2}

usually comes from mis-rationalizing the denominator as

\frac{\sqrt{3}-1}{2}

or mixing up the conjugate order; it is not equivalent to the original

x

. The choice

-\sqrt{3}

is double the correct result; it comes from simplifying

-\frac{2\sqrt{3}}{2}

-\sqrt{3}

instead of

-\sqrt{3}

, or from forgetting that rationalizing introduced the factor of

\frac{1}{2}

. Study tip: after rationalizing, write both

x^2

and

2x

with a common denominator before subtracting, and watch the negative sign carefully — sign errors are the most common trap on this type of problem.

Question 7

Which of the following is equal to $\sqrt{9+4\sqrt{5}}$ ?

$2-\sqrt{5}$
$3+2\sqrt{5}$
$\sqrt{5}-2$
$2+\sqrt{5}$ (correct answer)

Explanation: When you see a square root of a sum like

9+4\sqrt{5}

, the key is to recognize a perfect square hiding inside. Ask yourself: could this be the square of a two-term expression? Squaring

2+\sqrt{5}

gives

(2)^2+2(2)(\sqrt{5})+(\sqrt{5})^2=4+4\sqrt{5}+5=9+4\sqrt{5}

, so the principal square root is exactly

2+\sqrt{5}

. Now examine the traps. Both

2-\sqrt{5}

and

\sqrt{5}-2

square to

9-4\sqrt{5}

, the conjugate of your target, not the expression given. Additionally,

2-\sqrt{5}

is negative, so it cannot be a principal square root. The choice

3+2\sqrt{5}

might tempt you to just "split" the terms, but squaring it produces

29+12\sqrt{5}

, far too large. A reliable study strategy: when simplifying a nested radical, set up

(a+b\sqrt{c})^2

and match parts. Here the radical part is

4\sqrt{5}

, so the cross term

2ab\sqrt{5}

must equal

4\sqrt{5}

, meaning

ab=2

. Then look for integer values where

a^2+5b^2=9

; that quickly leads to

a=2, b=1

. Always verify by squaring your answer — it should bring you back to the original radicand exactly.

Question 8

For $x>0$ , simplify $\frac{5x}{\sqrt[3]{x}}$ .

$5x\sqrt[3]{x^2}$
$5\sqrt[3]{x}$
$5\sqrt[3]{x^2}$ (correct answer)
$5x^2$

Explanation: Whenever you see a radical like

\sqrt[3]{x}

, rewrite it as a fractional exponent:

\sqrt[3]{x}=x^{1/3}

. Then the expression becomes

\frac{5x}{x^{1/3}}

. Using the quotient rule for exponents,

x^{1}\div x^{1/3}=x^{1-1/3}=x^{2/3}

. So the simplification is

5x^{2/3}

, which converts back to

5\sqrt[3]{x^2}

. Now let's look at the traps. The choice

5x\sqrt[3]{x^2}

appears if you rationalize the denominator by multiplying by

\sqrt[3]{x^2}

but forget to cancel the

x

in the numerator with the

x

in the denominator — that product actually simplifies to

5\sqrt[3]{x^2}

. The choice

5\sqrt[3]{x}

would mean

5x^{1/3}

; this treats

x/\sqrt[3]{x}

as just

\sqrt[3]{x}

, but since

x=\sqrt[3]{x}\cdot\sqrt[3]{x}\cdot\sqrt[3]{x}

, dividing by

\sqrt[3]{x}

leaves two factors, not one. The choice

5x^2

comes from dropping the denominator in the fractional exponent:

x^{2/3}

is not

x^2

— the

2/3

means "cube root of

x^2

," not "x squared." Your go-to strategy: convert every radical to a fractional exponent first, combine exponents, and then convert back to radical form. This avoids both mis-applying exponent rules and miscounting cube-root factors.

Question 9

For $x>0$ , simplify $\sqrt[3]{x^2}\cdot\sqrt[4]{x^3}$ .

$x^{5/12}$
$x$
$x\sqrt[12]{x^5}$ (correct answer)
$x\sqrt[12]{x^7}$

Explanation: Whenever you see radicals with different indices, your first move should be to rewrite each in fractional-exponent form. Here,

\sqrt[3]{x^2}=x^{2/3}

and

\sqrt[4]{x^3}=x^{3/4}

, so the product is

x^{2/3}\cdot x^{3/4}

. Using the rule

x^a x^b=x^{a+b}

, add the exponents with a common denominator:

\frac{2}{3}+\frac{3}{4}=\frac{8}{12}+\frac{9}{12}=\frac{17}{12}

. So the simplified product is

x^{17/12}

. Split that into a whole

x

and a leftover root:

x^{17/12}=x\cdot x^{5/12}=x\sqrt[12]{x^5}

. That matches the correct choice. Now look at the traps. The choice

x^{5/12}

comes from adding the original numerators and multiplying the original denominators:

(2+3)/(3\cdot4)=5/12

. Fractional exponents must be added like regular fractions, not by this shortcut. The bare

x

is the result of pulling out

x^{12}

from

x^{17}

but forgetting the remaining

x^5

under the radical. Finally,

x\sqrt[12]{x^7}

would equal

x^{19/12}

; it appears if you slip on the addition and get

\frac{19}{12}

instead of

\frac{17}{12}

, or if you treat the remainder as

7

instead of

5

when splitting

x^{17}

. Study tip: on exponent/radical questions, convert to fractional exponents first, then use exponent laws. Double-check fraction addition with a common denominator—that single step will prevent most of these traps.

Question 10

For $x\ge0$ and $x\ne4$ , which expression is equivalent to $\frac{\sqrt{x}}{\sqrt{x}-2}$ ?

$\frac{x}{x-4}$
$\frac{x-2\sqrt{x}}{x-4}$
$\frac{x+2\sqrt{x}}{x+4}$
$\frac{x+2\sqrt{x}}{x-4}$ (correct answer)

Explanation: Whenever you see a radical in a denominator, rationalizing by multiplying by its conjugate is the key move. For

\sqrt{x}-2

, the conjugate is

\sqrt{x}+2

. Multiply both numerator and denominator by this same factor:

\frac{\sqrt{x}}{\sqrt{x}-2}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}. Since \((\sqrt{x}-2)(\sqrt{x}+2)=x-4\), and numerator becomes \(x+2\sqrt{x}\), the equivalent expression is \[ \frac{x+2\sqrt{x}}{x-4}.

The choice

\frac{x}{x-4}

is missing the

2\sqrt{x}

term: it treats

\sqrt{x}(\sqrt{x}+2)

as simply

x

, dropping the second product when distributing. The choice

\frac{x-2\sqrt{x}}{x-4}

has the correct denominator but the wrong sign in the numerator:it uses

\sqrt{x}-2

as the multiplier instead of the conjugate

\sqrt{x}+2

. The choice

\frac{x+2\sqrt{x}}{x+4}

has the correct numerator but the denominator wrong:

(a-b)(a+b)=a^2-b^2

, not

a^2+b^2

, so with

a=\sqrt{x}

and

b=2

you get

x-4

, not

x+4

. Remember: when rationalizing, write the conjugate with the opposite middle sign, then distribute carefully and verify that you multiplied the numerator and denominator by the exact same factor. This pattern will keep the signs and cross-terms straight.

Question 11

Simplify $\sqrt[4]{81x^8 y^{12}}$ . Assume all variables represent real numbers.

$3x^2 y^3$
$3x^2 |y^3|$ (correct answer)
$9x^2 |y^3|$
$3x^4 y^3$

Explanation: When you see a fourth root with variables and an even index, your first thought should be about absolute values: an even root of a variable expression must return a non-negative result for all real numbers. So you factor perfect fourth powers:

81 = 3^4

x^8 = (x^2)^4

, and

y^{12} = (y^3)^4

. Taking the fourth root gives

3

x^2

, and

|y^3|

— the absolute value is needed because

y^3

could be negative. Thus the simplified form is

3x^2 |y^3|

. Now the traps: The choice

3x^2 y^3

drops the absolute value on

y^3

, assuming

y

is non-negative — not allowed. The choice

9x^2 |y^3|

mistakenly takes the square root of 81 (giving 9) instead of the fourth root (giving 3). And

3x^4 y^3

divides the exponent 8 by 2 instead of 4, producing

x^4

instead of

x^2

, and also omits the absolute value. Your strategy: for even-index radicals, always check if any variable factor inside could be negative — if the exponent after dividing by the index is odd, wrap that factor in absolute value. Also, simplify the coefficient by finding the correct root, not the square root. This pattern appears often on algebra exams, so practice it until it becomes automatic.

Question 12

Simplify $3\sqrt{20} - \frac{10}{\sqrt{5}}$ .

$4\sqrt{5}$ (correct answer)
$8\sqrt{5}$
$-4\sqrt{5}$
$2\sqrt{5}$

Explanation: When you see a radical expression with a fraction, your instinct should be to simplify every radical and rationalize any denominator before combining. Start with

3\sqrt{20}=3\sqrt{4\cdot 5}=3\cdot 2\sqrt{5}=6\sqrt{5}.

Then handle the fraction:

\frac{10}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}}=\frac{10\sqrt{5}}{5}=2\sqrt{5}.

Now subtract like radicals:

6\sqrt{5}-2\sqrt{5}=4\sqrt{5},

4\sqrt{5}

is correct. The choice

8\sqrt{5}

comes from adding

6\sqrt{5}

and

2\sqrt{5}

, but the original operation is subtraction—preserve that minus sign. The choice

-4\sqrt{5}

often results from rationalizing incorrectly as

10\sqrt{5}

(forgetting to divide by 5) and then computing

6\sqrt{5}-10\sqrt{5}

, or from reversing the order of subtraction. The choice

2\sqrt{5}

is what you get if you simplify only the fraction and forget to simplify

3\sqrt{20}

first; you must simplify both terms before combining. A useful check: estimate with decimals. Since

6\sqrt{5}\approx 13.4

and

2\sqrt{5}\approx 4.5

, the difference is about 8.9, which matches

4\sqrt{5}\approx 8.9

. Always simplify and rationalize first, then combine like radicals using the sign exactly as written.

Question 13

A student rewrites $\frac{4}{\sqrt[3]{2}}$ by multiplying by $\frac{\sqrt[3]{2}}{\sqrt[3]{2}}$ and concludes that the simplified form is $2\sqrt[3]{2}$ . Which statement correctly identifies the error?

The student used the wrong multiplier; multiplying by $\frac{\sqrt[3]{4}}{\sqrt[3]{4}}$ gives $2\sqrt[3]{4}$ . (correct answer)
The student used the wrong multiplier; multiplying by $\frac{\sqrt[3]{4}}{\sqrt[3]{4}}$ gives $4\sqrt[3]{4}$ .
The student's method is correct because $\sqrt[3]{2}\cdot\sqrt[3]{2}=2$ , so the fraction can be reduced to $2\sqrt[3]{2}$ .
The expression is already simplified, since $\sqrt[3]{2}$ has no perfect-cube factor and an odd-index denominator cannot be rationalized.

Explanation: Clearing a cube-root denominator requires a multiplier that makes the radicand a perfect cube. Since

\sqrt[3]{2}\sqrt[3]{4}=\sqrt[3]{8}=2

, the correct multiplier is

\sqrt[3]{4}

, giving

\frac{4\sqrt[3]{4}}{2}=2\sqrt[3]{4}

. The student instead treated

\sqrt[3]{2}\cdot\sqrt[3]{2}

as 2 and did not actually clear the radical.

Question 14

For $x>0$ , which expression is equivalent to $\sqrt{x\sqrt{x\sqrt{x}}}$ ?

$x^{7/4}$
$x^{15/16}$
$x^{1/2}$
$x^{7/8}$ (correct answer)

Explanation: Rewrite each root as a power:

\sqrt{x\sqrt{x}}=x^{3/4}

, so the radicand is

x\cdot x^{3/4}=x^{7/4}

. Taking the outer square root gives

x^{7/8}

. The other choices come from omitting the outer root or combining the exponents incorrectly.

Question 15

For $x<3$ , which value is equivalent to $\frac{\sqrt{x^2-6x+9}}{x-3}$ ?

$x-3$
$-1$ (correct answer)
$1$
$3-x$

Explanation: Since

x^2-6x+9=(x-3)^2

, the numerator is

\sqrt{(x-3)^2}=|x-3|

. For

x<3

, this is

3-x

, while the denominator is

x-3=-(3-x)

. Therefore the quotient is

-1

. The choice

1

comes from canceling before applying the absolute value.

Question 16

If $x=\sqrt{7+4\sqrt3}$ , what is the value of $x+\frac{1}{x}$ ?

$2\sqrt3$
$4$ (correct answer)
$2+\sqrt3$
$2$

Explanation: Since

7+4\sqrt3=(2+\sqrt3)^2

, we have

x=2+\sqrt3

. Then

\frac{1}{x}=\frac{1}{2+\sqrt3}=2-\sqrt3

, so

x+\frac{1}{x}=4

. The choice

2+\sqrt3

is only the value of

x

, not the requested sum.

Question 17

For $x>0$ , $y>0$ , and $x\ne y$ , which expression is equivalent to $\frac{x-y}{\sqrt{x}+\sqrt{y}}+\frac{x-y}{\sqrt{x}-\sqrt{y}}$ ?

$2\sqrt{y}$
$0$
$2\sqrt{x}$ (correct answer)
$2\sqrt{x}-2\sqrt{y}$

Explanation: Each quotient is a difference of squares:

\frac{x-y}{\sqrt{x}+\sqrt{y}}=\sqrt{x}-\sqrt{y}

and

\frac{x-y}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}

. Their sum is

2\sqrt{x}

. The choice

2\sqrt{y}

results from confusing which variable remains.

Question 18

For all real $x$ , how many of the following simplifications are correct? I. $\sqrt{25x^2}=5x$ II. $\sqrt{8x^6}=2|x^3|\sqrt2$ III. $\sqrt[3]{27x^3}=3x$ IV. $\sqrt[4]{16x^8}=2x^2$

1
2
3 (correct answer)
4

Explanation: Statement I is not correct for negative

x

, because

\sqrt{25x^2}=5|x|

. Statements II, III, and IV are correct:

\sqrt{8x^6}=2|x^3|\sqrt2

, the cube root of a cube is

3x

for every real

x

, and the fourth root of

16x^8

2x^2

because

x^8

is nonnegative. Thus exactly 3 are correct.

Question 19

For all $x>0$ , if $\frac{k}{\sqrt{x+2}-\sqrt{x}}=2(\sqrt{x+2}+\sqrt{x})$ , what is the value of $k$ ?

2
4 (correct answer)
6
8

Explanation: Multiply the denominator by its conjugate:

(\sqrt{x+2}-\sqrt{x})(\sqrt{x+2}+\sqrt{x})=(x+2)-x=2

. Thus the left side is

\frac{k(\sqrt{x+2}+\sqrt{x})}{2}

, so

\frac{k}{2}=2

and

k=4

. The choice 2 comes from forgetting that the conjugate product is 2 rather than 1.

Question 20

Which expression is equivalent to $\frac{1}{\sqrt[3]{2}+\sqrt[3]{4}}$ after rationalizing the denominator?

$\frac{\sqrt[3]{4}+2-2\sqrt[3]{2}}{6}$
$\frac{1}{6}$
$\frac{2+\sqrt[3]{4}+\sqrt[3]{2}}{6}$
$\frac{\sqrt[3]{4}-2+2\sqrt[3]{2}}{6}$ (correct answer)

Explanation: Let

a=\sqrt[3]{2}

and

b=\sqrt[3]{4}=a^2

. Since

(a+b)(a^2-ab+b^2)=a^3+b^3=6

, multiply by

\frac{a^2-ab+b^2}{a^2-ab+b^2}

. The numerator is

b-2+2\sqrt[3]{2}

, giving

\frac{\sqrt[3]{4}-2+2\sqrt[3]{2}}{6}

. A sign error in the middle term is the main trap.