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Algebra II : Adding and Subtracting Radicals

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Example Question #51 : Adding And Subtracting Radicals

Add the radicals, if possible: $\sqrt{18}+\sqrt{50}+\sqrt{200}$

Possible Answers:

$18 \sqrt{2}$

$9 \sqrt{3}$

$\textup{The radicals cannot be added.}$

$6\sqrt3$

$24 \sqrt{2}$

Correct answer:

$18 \sqrt{2}$

Explanation:

Simplify all the radicals to their simplest forms. Use the perfect squares as the factors.

$\sqrt{18} =\sqrt{9\times 2}= \sqrt{9}\cdot \sqrt{2} = 3\sqrt{2}$

$\sqrt{50}= \sqrt{25}\cdot \sqrt{2} = 5 \sqrt{2}$

$\sqrt{200} = \sqrt{100}\cdot \sqrt{2} = 10\sqrt2$

Add the like terms together.

$3 \sqrt{2}+5 \sqrt{2}+10 \sqrt{2} = 18 \sqrt{2}$

The answer is: $18 \sqrt{2}$

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Example Question #91 : Simplifying Radicals

Simplify: $\sqrt{4x^2}-\sqrt{16y^4}$

Possible Answers:

2x^2-8y^2

2x-16y^2

2x-4y^2

12x-8y^2

2x-8y

Correct answer:

2x-4y^2

Explanation:

Simplify each radical first.

$\sqrt{4x^2}=2x$

$\sqrt{16y^4}=4y^2$

Now, subtract those:

2x-4y^2

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Example Question #53 : Adding And Subtracting Radicals

Add the radicals, if possible: $\sqrt8+\sqrt{20}+\sqrt{50}$

Possible Answers:

$\textup{The terms cannot be simplified.}$

$7\sqrt2+2\sqrt5$

$7\sqrt3+2\sqrt6$

$7\sqrt5+2\sqrt2$

Correct answer:

$7\sqrt2+2\sqrt5$

Explanation:

Rewrite the radicals using common factors of perfect squares.

$\sqrt8+\sqrt{20}+\sqrt{50} = \sqrt{4\times2}+\sqrt{4\times 5}+\sqrt{25\times 2}$

The equation becomes:

$\sqrt{4\times2}+\sqrt{4\times 5}+\sqrt{25\times 2}= 2\sqrt2+2\sqrt5 + 5\sqrt2$

Combine like-terms.

The answer is: $7\sqrt2+2\sqrt5$

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Example Question #101 : Simplifying Radicals

Add the radicals, if possible: $2\sqrt{27}+4\sqrt{24}+3\sqrt{12}$

Possible Answers:

$\textup{The radicals are already simplified.}$

$6\sqrt3+9\sqrt6$

$5\sqrt3+2\sqrt6$

$4\sqrt3+9\sqrt6$

$12\sqrt3+8\sqrt6$

Correct answer:

$12\sqrt3+8\sqrt6$

Explanation:

Evaluate each square root by factoring each with factors of perfect squares.

$\sqrt{27}= \sqrt{9}\cdot \sqrt 3 = 3\sqrt3$

$\sqrt{24} = \sqrt{4}\cdot \sqrt 6 = 2\sqrt6$

$\sqrt{12} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt3$

Replace all the terms in the expression.

$2\sqrt{27}+4\sqrt{24}+3\sqrt{12} = 2(3\sqrt3)+4(2\sqrt6)+3(2\sqrt{3})$

$6\sqrt3+8\sqrt6+6\sqrt3 = 12\sqrt3+8\sqrt6$

The answer is: $12\sqrt3+8\sqrt6$

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Example Question #55 : Adding And Subtracting Radicals

Add the radicals, if possible: $2\sqrt3+\sqrt6 +\sqrt{12}$

Possible Answers:

$2\sqrt6 +2\sqrt3$

$4\sqrt3 +\sqrt6$

$\textup{The radicals are already simplified.}$

$2\sqrt2 +2\sqrt3$

$8\sqrt2 +\sqrt6$

Correct answer:

$4\sqrt3 +\sqrt6$

Explanation:

Every radical in this expression is simplified except $\sqrt{12}$ .

Simplify by rewriting this radical using factors of perfect squares.

$\sqrt{12} = \sqrt{4}\cdot \sqrt{3} = 2\sqrt3$

Replace the term.

$2\sqrt3+\sqrt6 +\sqrt{12}=2\sqrt3+\sqrt6 +2\sqrt3$

Combine like-terms.

The answer is: $4\sqrt3 +\sqrt6$

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Example Question #56 : Adding And Subtracting Radicals

Simplify the expression:

$\sqrt{72} + \sqrt{98} - \sqrt{2}$

Possible Answers:

$4 \sqrt{21}$

$2\sqrt{42}$

$6\sqrt{14}$

$13\sqrt{2}$

$12\sqrt{2}$

Correct answer:

$12\sqrt{2}$

Explanation:

Simplify each of the three square root terms separately. Simplify $\sqrt{72}$ first as follows:

Express radicand 72 as the product of its prime factors:

$72 = 2 \cdot 36$

$= 2 \cdot 2 \cdot 18$

$= 2 \cdot 2 \cdot 2 \cdot 9$

$= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3$

Look for any prime factors that appear twice; there are two, 2 and 3, so restate the radical as

$\sqrt{72}$

$= \sqrt{ 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3}$

$= \sqrt{ 2 ^{2} \cdot 3 ^{2} \cdot 2}$

By the Product of Radicals Property, we can restate this as

$\sqrt{ 2 ^{2} } \cdot \sqrt{ 3 ^{2} } \cdot \sqrt{ 2}$

$=2 \cdot 3 \cdot \sqrt{ 2}$

$=6 \sqrt{ 2}$

The second term, $\sqrt{98}$ , can be simplified similarly:

$98 = 2 \cdot 49$

$= 2 \cdot 7 \cdot 7$

$\sqrt{98}$

$= \sqrt{2 \cdot 7 \cdot 7}$

$= \sqrt{ 7 ^{2}\cdot 2 }$

$= \sqrt{ 7 ^{2} } \cdot \sqrt{2}$

$= 7\sqrt{2}$

The third term, $\sqrt{2}$ , is already simplified, as 2 is prime.

Therefore,

$\sqrt{72} + \sqrt{98} - \sqrt{2}$

can be rewritten, and simplified using distribution:

$6 \sqrt{2} + 7 \sqrt{2} -1 \sqrt{2}$

$=( 6 + 7 -1 )\sqrt{2}$

$= 12 \sqrt{2}$

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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 #51 : Adding And Subtracting Radicals

Example Question #91 : Simplifying Radicals

Example Question #53 : Adding And Subtracting Radicals

Example Question #101 : Simplifying Radicals

Example Question #55 : Adding And Subtracting Radicals

Example Question #56 : Adding And Subtracting Radicals

All Algebra II Resources

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