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High School Math : Radicals

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Example Questions

Example Question #31 : Radicals

Simplify the following radical expression:

$\sqrt[3]{(3x-y)^5}$

Possible Answers:

$(3x-y)\sqrt[3]{(3x-y)^2}$

$(3x-y)^2\sqrt[3]{(3x-y)^3}$

$(3x-y)\sqrt[3]{(3x-y)^3}$

$(3x-y)\sqrt[3]{(3x-y)}$

$(3x-y)^2\sqrt[3]{(3x-y)^2}$

Correct answer:

$(3x-y)\sqrt[3]{(3x-y)^2}$

Explanation:

Simplify the radical expression:

$\sqrt[3]{(3x-y)^5}$

$\sqrt[3]{(3x-y)^2(3x-y)^3}$

$(3x-y)\sqrt[3]{(3x-y)^2}$

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Example Question #31 : Radicals

Simplify the expression:

$\frac{3\sqrt[4]{32}}{2\sqrt[4]{162}}$

Possible Answers:

. $\frac{3\sqrt[4]{16}}{2\sqrt[4]{81}}$

$\frac{2}{3}$

$6\sqrt[4]{2}$

$\frac{3}{2}$

Correct answer:

Explanation:

Use the multiplication property of radicals to split the fourth roots as follows:

$\rightarrow \frac{3\sqrt[4]{16}\sqrt[4]{2}}{2\sqrt[4]{81}\sqrt[4]{2}}$

Simplify the new roots:

$\rightarrow \frac{3(2)\sqrt[4]{2}}{2(3)\sqrt[4]{2}}$

$\rightarrow \frac{6\sqrt[4]{2}}{6\sqrt[4]{2}}$

$\rightarrow 1$

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Example Question #1 : Radicals

Find the value of $\sqrt{80} + \sqrt{20}$ .

Possible Answers:

$4\sqrt5$

$\sqrt5$

$6\sqrt5$

$6\sqrt{10}$

Correct answer:

$6\sqrt5$

Explanation:

To solve this equation, we have to factor our radicals. We do this by finding numbers that multiply to give us the number within the radical.

$\sqrt{80} = \sqrt{20}\sqrt{4}$

$\sqrt{20} = \sqrt{5}\sqrt{4}$

Add them together:

$\sqrt5\sqrt4\sqrt4 + \sqrt5\sqrt4$

4 is a perfect square, so we can find the root:

$(2)(2)\sqrt5 + 2\sqrt5$

$4\sqrt5 + 2\sqrt5 = 6\sqrt5$

Since both have the same radical, we can combine them:

$4\sqrt5 + 2\sqrt5 = 6\sqrt5$

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Example Question #12 : Simplifying Radicals

Factor and simplify the following radical expression:

$(4-2\sqrt{2})(\sqrt{2}-4)$

Possible Answers:

$12\sqrt{2}-18$

$10\sqrt{2}-20$

$12\sqrt{2}-20$

$14\sqrt{2}-20$

$12\sqrt{2}-16$

Correct answer:

$12\sqrt{2}-20$

Explanation:

Begin by using the FOIL method (First Outer Inner Last) to expand the expression.

$(4-2\sqrt{2})(\sqrt{2}-4)$

$4\sqrt{2}-16-4+8\sqrt{2}$

Now, combine like terms:

$12\sqrt{2}-20$

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Example Question #13 : Simplifying Radicals

What is the value of $5\sqrt5 + 3 - 4\sqrt{2} +9 - 3\sqrt{5}$ ?

Possible Answers:

$4\sqrt{7} - 12$

$-2\sqrt{3} + 12$

$2\sqrt{5} - 4\sqrt{2} + 12$

$2\sqrt{10} - 4\sqrt{2} + 12$

$2\sqrt{7} - 4\sqrt{7} + 12$

Correct answer:

$2\sqrt{5} - 4\sqrt{2} + 12$

Explanation:

When combining terms involving radicals, we can only combine the ones that have the same radical. For this problem, that means $4\sqrt2$ has to stay on its own while we can combine $5\sqrt5$ and $-3\sqrt5$ into $2\sqrt5$ . The simple integers can be combined too, giving us our answer with three seperate terms.

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Example Question #32 : Radicals

Simplify the expression:.

$\frac{x^{\frac{5}{6}}}{x^{\frac{7}{12}}}$

Possible Answers:

$\sqrt{x^4}$

$\sqrt[4]{x}$

$\sqrt[6]{x^2}$

$\sqrt{x^6}$

$\sqrt[6]{x}$

Correct answer:

$\sqrt[4]{x}$

Explanation:

Exponents in the denominator can be subtracted from exponents in the numerator.

$\small \frac{x^{\frac{5}{6}}}{x^{\frac{7}{12}}}=x^{\frac{5}{6}-\frac{7}{12}}=x^{\frac{10}{12}-\frac{7}{12}}=x^{\frac{3}{12}}=x^{\frac{1}{4}}$

Recall that $x^\frac{a}{b}=\sqrt[b]{x^a}$ .

Therefore, $x^\frac{1}{4}=\sqrt[4]{x}$ .

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Example Question #33 : Radicals

Simplify:

$\sqrt{75}-\sqrt{27}$

Possible Answers:

$3\sqrt{5}$

$3\sqrt{2}$

$5\sqrt{3}$

$5\sqrt{2}$

$2\sqrt{3}$

Correct answer:

$2\sqrt{3}$

Explanation:

Try to group factors in pairs to get perfect squares under the square root:

$\sqrt{75}-\sqrt{27}= \sqrt{25\cdot 3}-\sqrt{9\cdot 3}=5\sqrt{3}-3\sqrt{3}=2\sqrt{3}$

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Example Question #1 : Solving Radical Equations And Inequalities

Solve for :

$\sqrt{x^2+12x+36} =4$

Possible Answers:

$x = \left \{ -10, -2\right \}$

x = -2

$x = \{-10, -6, -2\}$

$x = \{-2, 0\}$

x = -10

Correct answer:

$x = \left \{ -10, -2\right \}$

Explanation:

To solve for in the equation $\sqrt{x^2+12x+36} =4$

Square both sides of the equation

$\rightarrow (\sqrt{x^2+12x+36})^2 =(4)^2$

$\rightarrow x^2+12x+36 =16$

Set the equation equal to by subtracting the constant from both sides of the equation.

$\rightarrow (x^2+12x+36) - 16 =(16) -16$

$\rightarrow x^2+12x+20 =0$

Factor to find the zeros:

$\rightarrow (x+10)(x+2) = 0$

This gives the solutions

$x = \left \{ -10, -2\right \}$ .

Verify that these work in the original equation by substituting them in for . This is especially important to do in equations involving radicals to ensure no imaginary numbers (square roots of negative numbers) are created.

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Example Question #2 : Solving Radical Equations And Inequalities

Solve the following radical expression:

$\sqrt{3x+1}+2=7$

Possible Answers:

Correct answer:

Explanation:

Begin by subtracting from each side of the equation:

$\sqrt{3x+1}+2=7$

$\sqrt{3x+1}=5$

Now, square the equation:

$(\sqrt{3x+1}=5)^2$

3x+1=25

Solve the linear equation:

3x=24

x=8

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Example Question #3 : Solving And Graphing Radical Equations

Solve the following radical expression:

$\sqrt{2x+1}-2=\sqrt{2x-7}$

Possible Answers:

Correct answer:

Explanation:

Begin by squaring both sides of the equation:

$\sqrt{2x+1}-2=\sqrt{2x-7}$

$(\sqrt{2x+1}-2)^2=(\sqrt{2x-7})^2$

$(2x+1)-4\sqrt{2x+1}+4=2x-7$

Combine like terms:

$-4\sqrt{2x+1}=-12$

$\sqrt{2x+1}=3$

Once again, square both sides of the equation:

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

2x+1=9

Solve the linear equation:

2x=8

x=4

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High School Math : Radicals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #31 : Radicals

Example Question #31 : Radicals

Example Question #1 : Radicals

Example Question #12 : Simplifying Radicals

Example Question #13 : Simplifying Radicals

Example Question #32 : Radicals

Example Question #33 : Radicals

Example Question #1 : Solving Radical Equations And Inequalities

Example Question #2 : Solving Radical Equations And Inequalities

Example Question #3 : Solving And Graphing Radical Equations

All High School Math Resources

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