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Algebra II : Adding and Subtracting Radicals

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Example Questions

Example Question #1 : Radicals

Subtract and simplify:
$\small \frac{\sqrt3}{\sqrt5}-\frac{\sqrt3}{\sqrt2}$

Possible Answers:

$\small \frac{\sqrt{60}-\sqrt{150}}{10}$

$\small \frac{2\sqrt{15}-5\sqrt6}{10}$

$\small 3(\sqrt2-\sqrt5)$

$\small \frac{\sqrt{6}-\sqrt{15}}{\sqrt{10}}$

Correct answer:

$\small \frac{2\sqrt{15}-5\sqrt6}{10}$

Explanation:

$\small \frac{\sqrt3}{\sqrt5}-\frac{\sqrt3}{\sqrt2}$

Find the lease common denominator: $\small \sqrt{10}$

$\small \small \frac{\sqrt{3}}{\sqrt5}=\frac{\sqrt6}{\sqrt{10}}$

$\small \small \frac{\sqrt{3}}{\sqrt2}=\frac{\sqrt{15}}{\sqrt{10}}$

$\small \small \frac{\sqrt6}{\sqrt10}-\frac{\sqrt{15}}{\sqrt{10}}=\frac{\sqrt6-\sqrt{15}}{\sqrt{10}}$

A radical cannot be in the denominator:

$\small \small (\frac{\sqrt6-\sqrt{15}}{\sqrt{10}})(\frac{\sqrt{10}}{\sqrt{10}})=\frac{\sqrt{60}-\sqrt{150}}{10}$

$\small \sqrt{60}=2\sqrt{15}$

$\small \sqrt{150}=5\sqrt6$

$\small \frac{\sqrt{60}-\sqrt{150}}{10}=\frac{2\sqrt{15}-5\sqrt6}{10}$

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Example Question #1 : Radicals

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

Possible Answers:

$6\sqrt{10}$

$6\sqrt5$

$4\sqrt5$

$\sqrt5$

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 #2 : Radicals

Simplify the following equation:

$f(x)=x^{2}-3\sqrt{4x}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5\sqrt[3]{x}$

Possible Answers:

Cannot simplify further

$f(x)=x^{2}-x-13\sqrt{x}+9\sqrt[3]{x}$

$f(x)=x^{2}-x-4x^{\frac{1}{6}}$

$f(x)=x^{2}-x+13\sqrt{x}-9\sqrt[3]{x}$

Correct answer:

$f(x)=x^{2}-x-13\sqrt{x}+9\sqrt[3]{x}$

Explanation:

When simplifying, you should always be on the lookout for like terms. While it might not look like there are like terms in f(x) , there are -- we just have to be able to rewrite it to see.

$f(x)=x^{2}-3\sqrt{4x}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5\sqrt[3]{x}$

$f(x)=x^{2}-3(4x)^{\frac{1}{2}}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5x^{\frac{1}{3}}$

Before we start combining terms, though, let's look a little more closely at this part:

$-3(4x)^{\frac{1}{2}}$

We need to "distribute" that exponent to everything in the parentheses, like so:

$-3\cdot 4^{\frac{1}{2}}\cdot x^{\frac{1}{2}}$

But 4 to the one-half power is just the square root of 4, or 2.

$-3\cdot 2\cdot x^{\frac{1}{2}}=-6x^{\frac{1}{2}}$

Okay, now let's see our equation.

$f(x)=x^{2}-6x^{\frac{1}{2}}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5x^{\frac{1}{3}}$

We need to start combining like terms. Take the terms that include x to the one-half power first.

$f(x)=x^{2}\mathbf{-6x^{\frac{1}{2}}}+4x^{\frac{1}{3}}-x\mathbf{-7x^{\frac{1}{2}}}+5x^{\frac{1}{3}}=x^{2}\mathbf{-13x^{\frac{1}{2}}}+4x^{\frac{1}{3}}-x+5x^{\frac{1}{3}}$

Now take the terms that have x to the one-third power.

$f(x)=x^{2}-13x^{\frac{1}{2}}\mathbf{+4x^{\frac{1}{3}}}-x\mathbf{+5x^{\frac{1}{3}}}=x^{2}-13x^{\frac{1}{2}}\mathbf{+9x^{\frac{1}{3}}}-x$

All that's left is to write them in order of descending exponents, then convert the fractional exponents into radicals (since that's what our answer choices look like).

$f(x)=x^{2}-x-13x^{\frac{1}{2}}+9x^{\frac{1}{3}}$

$f(x)=x^{2}-x-13\sqrt{x}+9\sqrt[3]{x}$

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Example Question #3 : Adding And Subtracting Radicals

Simplify.

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

Possible Answers:

$8-4\sqrt{8}$

$8-8{\sqrt{2}}$

$8-4\sqrt{2}$

${\sqrt{2}}$

Correct answer:

$8-8{\sqrt{2}}$

Explanation:

We can simplify both radicals:

$\sqrt{16} =4$ and $\sqrt{8}=\sqrt{4\times 2} = 2\sqrt{2}$

Plug in the simplified radicals into the equation:

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

which leaves us with:

$8- 8\sqrt{2}$

Because these are not like terms, we cannot simplify this further.

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Example Question #1 : Adding And Subtracting Radicals

Simplify.

$-8\sqrt{27}-11\sqrt{9}+2\sqrt{3}$

Possible Answers:

$-22\sqrt{3}+33$

$-22\sqrt{3}-33$

$22\sqrt{3}+33$

$-26\sqrt{3}-33$

Correct answer:

$-22\sqrt{3}-33$

Explanation:

Only the first two radicals can be simplified:

$\sqrt{27}=\sqrt{9\times 3}=3\sqrt{3}$ and $\sqrt{9}=3$

Plug in the simplified radicals into the equation:

$-8(3\sqrt{3})-11(3)+2\sqrt{3}$

$-24\sqrt{3}-33+2\sqrt{3}$

We can now simplify the equation by combining the like terms:

$-22\sqrt{3}-33$

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Example Question #2 : Adding And Subtracting Radicals

Simplify.

$5\sqrt{28} +\sqrt{7} -7\sqrt{63}$

Possible Answers:

$-10\sqrt7{}$

$-11\sqrt{7}$

$10\sqrt{7}-21\sqrt{63}$

$6\sqrt{7}-7\sqrt{63}$

Correct answer:

$-10\sqrt7{}$

Explanation:

We can simplify the first and third radicals:

$\sqrt{28} =\sqrt{7\times 4} =2\sqrt{7}$ and $\sqrt{63} =\sqrt{9\times 7 } =3\sqrt{7}$

Plug in the simplified radicals into the equation:

$5(2\sqrt{7}) + \sqrt{7} -7(3\sqrt{7})$

$10\sqrt{7} + \sqrt{7 } -21\sqrt{7}$

Combine the like terms:

$-10\sqrt{7}$

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Example Question #1 : Adding And Subtracting Radicals

$What\: is \; 3\sqrt{7} + 4\sqrt{7}\, ?$

Possible Answers:

$12\sqrt7$

$7\sqrt7$

$7\sqrt14$

Correct answer:

$7\sqrt7$

Explanation:

$Think \: of\; x = \sqrt7$

3x + 4x = 7x

$Therefore, \: 3\sqrt{7} + 4\sqrt{7} = 7\sqrt7$

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Example Question #2552 : Algebra 1

$\sqrt[3]{27} + \sqrt{64} =$

Possible Answers:

Correct answer:

Explanation:

The third root of is

$\left ( 3\times 3\times 3 =27\right )$

and when added to the square root of 64, which is 8, you should get 11.

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Example Question #4041 : Algebra Ii

Solve.

${}\sqrt{2}+\sqrt{3}$

Possible Answers:

$\sqrt{6}$

$\sqrt{5}$

${}\sqrt{2}+\sqrt{3}$

$\sqrt{10}$

Correct answer:

${}\sqrt{2}+\sqrt{3}$

Explanation:

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical.

Since they are not the same, the answer is just the problem stated.

${}\sqrt{2}+\sqrt{3}$

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Example Question #51 : Simplifying Radicals

Solve.

$\sqrt{7}-\sqrt{4}$

Possible Answers:

$\sqrt{6}$

$\sqrt{28}$

$\sqrt{11}$

${}\sqrt{7}-\sqrt{4}$

$\sqrt{3}$

Correct answer:

${}\sqrt{7}-\sqrt{4}$

Explanation:

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical.

Since they are not the same, the answer is just the problem stated.

$\sqrt{7}-\sqrt{4}$

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Algebra II : Adding and Subtracting Radicals

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Radicals

Example Question #1 : Radicals

Example Question #2 : Radicals

Example Question #3 : Adding And Subtracting Radicals

Example Question #1 : Adding And Subtracting Radicals

Example Question #2 : Adding And Subtracting Radicals

Example Question #1 : Adding And Subtracting Radicals

Example Question #2552 : Algebra 1

Example Question #4041 : Algebra Ii

Example Question #51 : Simplifying Radicals

All Algebra II Resources

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