College Algebra Quiz: Simplifying Radicals And Combining Like Radicals

4 questions · exam conditions

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Simplifying Radicals And Combining Like RadicalsQuestion 1 of 4

Rationalize the denominator and simplify: $\frac{4}{\sqrt{12} + \sqrt{3}}$

\frac{4\sqrt{3}}{9}

\frac{4(\sqrt{12} - \sqrt{3})}{9}

\frac{4\sqrt{3}}{3}

\frac{4}{3\sqrt{3}}

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College Algebra Quiz

College Algebra Quiz: Simplifying Radicals And Combining Like Radicals

Practice Simplifying Radicals And Combining Like Radicals in College Algebra 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 Simplifying Radicals And Combining Like Radicals, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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

Rationalize the denominator and simplify: $\frac{4}{\sqrt{12} + \sqrt{3}}$

$\frac{4\sqrt{3}}{9}$ (correct answer)
$\frac{4(\sqrt{12} - \sqrt{3})}{9}$
$\frac{4\sqrt{3}}{3}$
$\frac{4}{3\sqrt{3}}$

Explanation: First simplify

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

. So the expression becomes

\frac{4}{2\sqrt{3} + \sqrt{3}} = \frac{4}{3\sqrt{3}}

. To rationalize, multiply by

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

\frac{4}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{3 \cdot 3} = \frac{4\sqrt{3}}{9}

. Choice B incorrectly uses the conjugate method on the simplified form. Choice C has the wrong denominator (should be 9, not 3). Choice D is the intermediate step before rationalization.

Question 2

Which expression is equivalent to $\sqrt{72x^3y^5}$ when simplified completely, assuming all variables represent positive real numbers?

$6xy^2\sqrt{2xy}$ (correct answer)
$6x^2y^3\sqrt{2x}$
$6xy^2\sqrt{2x}$
$36xy^2\sqrt{2x}$

Explanation: To simplify

\sqrt{72x^3y^5}

, first factor out perfect squares:

72 = 36 \cdot 2 = 6^2 \cdot 2

x^3 = x^2 \cdot x

, and

y^5 = y^4 \cdot y = (y^2)^2 \cdot y

. So

\sqrt{72x^3y^5} = \sqrt{6^2 \cdot 2 \cdot x^2 \cdot x \cdot (y^2)^2 \cdot y} = 6 \cdot x \cdot y^2 \sqrt{2xy} = 6xy^2\sqrt{2xy}

. Choice B incorrectly factors the exponents as

x^3 = x^2 \cdot x

but writes

x^2

outside. Choice C misses the

y

under the radical. Choice D incorrectly takes

\sqrt{36} = 36

Question 3

Simplify $\sqrt{8x^2} + x\sqrt{18} - 2\sqrt{2x^2}$ where $x > 0$ .

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

Explanation: Since

x > 0

, we have

\sqrt{x^2} = x

. Simplifying each term:

\sqrt{8x^2} = \sqrt{4 \cdot 2 \cdot x^2} = 2x\sqrt{2}

x\sqrt{18} = x\sqrt{9 \cdot 2} = 3x\sqrt{2}

, and

2\sqrt{2x^2} = 2x\sqrt{2}

. Combining like terms:

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

. Choice A results from incorrectly combining coefficients. Choice D shows the expression before combining like terms. Choice C adds all coefficients without considering the subtraction.

Question 4

If $\sqrt{a} + \sqrt{b} = 7$ and $\sqrt{a} - \sqrt{b} = 1$ , what is the value of $ab$ ?

$12$
$16$
$144$ (correct answer)
$49$

Explanation: Let

x = \sqrt{a}

and

y = \sqrt{b}

. Then we have the system:

x + y = 7

and

x - y = 1

. Adding these equations:

2x = 8

, so

x = 4

, which means

\sqrt{a} = 4

and

a = 16

. Substituting back:

4 + y = 7

, so

y = 3

, which means

\sqrt{b} = 3

and

b = 9

. Therefore,

ab = 16 \cdot 9 = 144

. Choice A represents

a + b

. Choice B represents just

a

. Choice D represents

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

without expanding correctly.