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Algebra II : Simplifying Radicals

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Example Questions

Example Question #52 : Adding And Subtracting Radicals

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

Possible Answers:

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

$7\sqrt3+2\sqrt6$

$7\sqrt5+2\sqrt2$

$7\sqrt2+2\sqrt5$

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:

$12\sqrt3+8\sqrt6$

$5\sqrt3+2\sqrt6$

$6\sqrt3+9\sqrt6$

$4\sqrt3+9\sqrt6$

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

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 #1433 : Mathematical Relationships And Basic Graphs

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

Possible Answers:

$2\sqrt2 +2\sqrt3$

$8\sqrt2 +\sqrt6$

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

$2\sqrt6 +2\sqrt3$

$4\sqrt3 +\sqrt6$

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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Example Question #1 : Radicals

Multiply and express the answer in the simplest form:

$\small \sqrt3(\sqrt4-\sqrt3)$

Possible Answers:

$\small 4\sqrt3-9$

$\small \sqrt7-\sqrt6$

$\small \sqrt{12}-\sqrt9$

$\small 2\sqrt3-3$

Correct answer:

$\small 2\sqrt3-3$

Explanation:

$\small \sqrt3(\sqrt4-\sqrt3)$

$\small \sqrt{12}-\sqrt9$

$\small \sqrt{12}=\sqrt4 \times \sqrt3=2\sqrt3$

$\small \sqrt9=3$

$\small \sqrt{12}-\sqrt9=2\sqrt3-3$

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Example Question #1 : Multiplying And Dividing Radicals

$\left ( \sqrt{x}+\sqrt{6} \right )\left ( \sqrt{x}-\sqrt{6} \right )$

Possible Answers:

x-6

$x-\sqrt{12}$

$x^{2}-36$

$x-\sqrt{6}$

$\sqrt{x}-6$

Correct answer:

x-6

Explanation:

FOIL with difference of squares. The multiplying cancels the square roots on both terms.

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Example Question #101 : Simplifying Radicals

Simplify.

$\sqrt{36} \times 3\sqrt{36}$

Possible Answers:

$3\sqrt{36}$

108

$4\sqrt{36}$

Correct answer:

108

Explanation:

We can solve this by simplifying the radicals first: $\sqrt{36} =6$

Plugging this into the equation gives us:

$6 \times3(6)=108$

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Example Question #2 : Multiplying And Dividing Radicals

Simplify.

$5\sqrt{2} \times4\sqrt{2}$

Possible Answers:

$20\sqrt{2}$

$\sqrt{80}$

Correct answer:

Explanation:

Note: the product of the radicals is the same as the radical of the product:

$\sqrt{2} \times \sqrt2} = \sqrt{2\times 2$ which is $\sqrt{4} =2$

Once we understand this, we can plug it into the equation:

$5\sqrt{2} \times4\sqrt{2} = 5\times4\times \sqrt{2\times 2}$

$5\times4\times2=40$

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Example Question #2 : Multiplying And Dividing Radicals

Simplify.

$\frac{\sqrt{32}}{3\sqrt{16}}$

Possible Answers:

$\frac{4}{12}$

$\frac{4\sqrt{2}}{12}$

$\frac{\sqrt{2}}{3}$

$\sqrt{2}$

Correct answer:

$\frac{\sqrt{2}}{3}$

Explanation:

We can simplify the radicals:

$\sqrt{32} =\sqrt{16\times 2} =4\sqrt{2}$ and $\sqrt{16} =4$

Plug in the simplifed radicals into the equation:

$\frac{4\sqrt{2}}{3\times 4} = \frac{4\sqrt{2}}{12}$

$\frac{4 \sqrt2}{12}=\frac{\sqrt2}{3}$

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Example Question #2 : Multiplying And Dividing Radicals

Simplify and rationalize the denominator if needed,

$\frac{\sqrt{16}}{4\sqrt{2}}$

Possible Answers:

$\frac{4}{\sqrt{2}}$

$\frac{1}{2}$

$\sqrt{2}$

$\frac{\sqrt2}{2}$

Correct answer:

$\frac{\sqrt2}{2}$

Explanation:

We can only simplify the radical in the numerator: $\sqrt{16} =4$

Plugging in the simplifed radical into the equation we get:

$\frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}}$

Note: We simplified further because both the numerator and denominator had a "4" which canceled out.

Now we want to rationalize the denominator,

$\frac{1}{\sqrt2}\bigg(\frac{\sqrt2}{\sqrt2}\bigg)=\frac{\sqrt2}{2}$

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Algebra II : Simplifying Radicals

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #52 : Adding And Subtracting Radicals

Example Question #101 : Simplifying Radicals

Example Question #1433 : Mathematical Relationships And Basic Graphs

Example Question #56 : Adding And Subtracting Radicals

Example Question #1 : Radicals

Example Question #1 : Multiplying And Dividing Radicals

Example Question #101 : Simplifying Radicals

Example Question #2 : Multiplying And Dividing Radicals

Example Question #2 : Multiplying And Dividing Radicals

Example Question #2 : Multiplying And Dividing Radicals

All Algebra II Resources

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