GRE Radicals Questions Practice Questions with Answers

Radicals represent the inverse operation of exponentiation, typically denoted by the symbol $\sqrt{}$, and are essential for solving geometry and algebra problems on the GRE. Understanding how to manipulate these roots is vital for success in the Quantitative Reasoning section, where you will frequently encounter square roots, cube roots, and higher-order radicals. These concepts are foundational for more complex topics found in GRE Prep resources.

Concept Explanation

A radical is a mathematical expression involving a root, where the nth root of a number $x$ is a value that, when multiplied by itself $n$ times, equals $x$. In the context of the GRE, the most common radical is the square root ($n=2$), but you may also see cube roots ($n=3$). Radicals can also be expressed as fractional exponents; for example, $\sqrt{x}$ is equivalent to $x^{1/2}$. Key properties you must know include the Product Rule, $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$, and the Quotient Rule, $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$. It is critical to remember that you can only add or subtract radicals if they have the same radicand (the number under the symbol), such as $3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$. For more advanced practice, tools like the AI Question Generator can help you create custom drills on these specific rules.

Solved Examples

1. Simplify the expression: $\sqrt{72} + \sqrt{50}$
   1. Factor each radicand into a perfect square and another factor: $72 = 36 \times 2$ and $50 = 25 \times 2$.
   2. Apply the product rule: $\sqrt{36 \times 2} + \sqrt{25 \times 2}$.
   3. Simplify the square roots of the perfect squares: $6\sqrt{2} + 5\sqrt{2}$.
   4. Combine like terms: $11\sqrt{2}$.
2. Solve for $x$: $\sqrt{x + 5} = 4$
   1. Square both sides of the equation to remove the radical: $(\sqrt{x + 5})^2 = 4^2$.
   2. Simplify: $x + 5 = 16$.
   3. Subtract 5 from both sides: $x = 11$.
3. Rationalize the denominator: $\frac{6}{\sqrt{3}}$
   1. Multiply both the numerator and denominator by $\sqrt{3}$ to eliminate the radical in the denominator: $\frac{6 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$.
   2. Simplify the denominator: $\frac{6\sqrt{3}}{3}$.
   3. Divide the coefficients: $2\sqrt{3}$.
4. Simplify $\sqrt[3]{54}$
   1. Identify the largest perfect cube factor of 54, which is 27 ($3^3$): $54 = 27 \times 2$.
   2. Apply the property: $\sqrt[3]{27} \times \sqrt[3]{2}$.
   3. Simplify the cube root: $3\sqrt[3]{2}$.

Practice Questions

1. Simplify the expression $\sqrt{48} - \sqrt{12}$.

2. If $y = \sqrt{144} + \sqrt{81}$, what is the value of $y$?

3. Simplify $\frac{\sqrt{75}}{\sqrt{3}}$.

4. Solve for $z$: $2\sqrt{z} = 10$.

5. Simplify $\sqrt{x^6 y^4}$, assuming $x$ and $y$ are positive.

6. Evaluate $(\sqrt{7} + \sqrt{2})(\sqrt{7} - \sqrt{2})$.

7. Simplify $3\sqrt{20} + \sqrt{45}$.

8. What is the value of $\sqrt[3]{-125}$?

9. Rationalize the denominator of $\frac{10}{\sqrt{5}}$.

10. Compare Quantity A and Quantity B:
Quantity A: $\sqrt{100 - 64}$
Quantity B: $\sqrt{100} - \sqrt{64}$

Answers & Explanations

1. $2\sqrt{3}$: Simplify $\sqrt{48}$ to $\sqrt{16 \times 3} = 4\sqrt{3}$ and $\sqrt{12}$ to $\sqrt{4 \times 3} = 2\sqrt{3}$. Then, $4\sqrt{3} - 2\sqrt{3} = 2\sqrt{3}$.
2. 21: $\sqrt{144} = 12$ and $\sqrt{81} = 9$. Therefore, $12 + 9 = 21$.
3. 5: Using the quotient rule, $\sqrt{\frac{75}{3}} = \sqrt{25} = 5$.
4. 25: Divide both sides by 2 to get $\sqrt{z} = 5$. Square both sides: $z = 25$.
5. $x^3 y^2$: Take the square root of each variable by dividing the exponent by 2: $x^{6/2} y^{4/2} = x^3 y^2$.
6. 5: This is a difference of squares: $(\sqrt{7})^2 - (\sqrt{2})^2 = 7 - 2 = 5$.
7. $9\sqrt{5}$: Simplify $3\sqrt{20} = 3\sqrt{4 \times 5} = 3 \times 2\sqrt{5} = 6\sqrt{5}$. Simplify $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$. Add them: $6\sqrt{5} + 3\sqrt{5} = 9\sqrt{5}$.
8. -5: The cube root of a negative number is negative. Since $(-5)^3 = -125$, the answer is -5.
9. $2\sqrt{5}$: Multiply numerator and denominator by $\sqrt{5}$ to get $\frac{10\sqrt{5}}{5}$, which simplifies to $2\sqrt{5}$.
10. Quantity A is greater: Quantity A is $\sqrt{36} = 6$. Quantity B is $10 - 8 = 2$. Since $6 > 2$, Quantity A is greater. For more strategy on comparison questions, see our study guide on effective practice.

Frequently Asked Questions

Can you add radicals with different numbers under the root?

No, you cannot directly add radicals like $\sqrt{2} + \sqrt{3}$ into a single radical term. You can only add or subtract "like radicals" that have the exact same radicand and index.

What does it mean to rationalize a denominator?

Rationalizing a denominator is the process of eliminating radical signs from the bottom of a fraction. This is typically done by multiplying the numerator and denominator by the radical that appears in the denominator.

Does the GRE allow calculators for radical calculations?

The GRE provides an on-screen calculator that has a square root function. However, the calculator does not simplify radicals into exact forms like $2\sqrt{3}$, so manual simplification skills are still necessary.

Are square roots on the GRE always positive?

When the radical symbol $\sqrt{}$ is used on the GRE, it refers specifically to the non-negative (principal) square root. For example, $\sqrt{25}$ is 5, not \u00b15.

How do fractional exponents relate to radicals?

Fractional exponents are an alternative way to write radicals where the denominator of the exponent represents the root index. For instance, $x^{1/3}$ is the cube root of $x$ and $x^{2/3}$ is the cube root of $x$ squared.