College Algebra Quiz: Simplifying Radicals And Combining Like Radicals

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.

Question 1

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

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 2

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

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 3

1. $12$
2. $16$
3. $144$ (correct answer)
4. $49$

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.

Question 4

1. $6xy^2\sqrt{2xy}$ (correct answer)
2. $6x^2y^3\sqrt{2x}$
3. $6xy^2\sqrt{2x}$
4. $36xy^2\sqrt{2x}$

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 $\sqrt{36} = 36$.