PSAT Math : How to multiply square roots

Study concepts, example questions & explanations for PSAT Math

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Example Questions

Example Question #1 : How To Multiply Square Roots

Multiply and simplify. Assuming all integers are positive real numbers. 

\(\displaystyle 2\sqrt{5}\cdot6\sqrt{20}\)

Possible Answers:

\(\displaystyle 160\)

\(\displaystyle 40\)

\(\displaystyle 60\)

\(\displaystyle 120\)

\(\displaystyle 80\)

Correct answer:

\(\displaystyle 120\)

Explanation:

\(\displaystyle 2\sqrt{5}\cdot6\sqrt{20}\)

Multiply the coefficents outside of the radicals.

\(\displaystyle 2\cdot6= 12\)

Then multiply the radicans. Simplify by checking for a perfect square. 

\(\displaystyle \sqrt{5\cdot 20}=\sqrt{100}=10\)

Final answer is your leading coefficent, \(\displaystyle 12\), multiplied by the answer acquired by multiplying the terms under the radican, \(\displaystyle 10\)

\(\displaystyle 12\cdot10= 120\)

The final answer is \(\displaystyle 120\).

Example Question #2 : How To Multiply Square Roots

Mulitply and simplify. Assume all integers are positive real numbers. 

\(\displaystyle \sqrt{3}(3\sqrt{2}+\sqrt{3})\)

Possible Answers:

\(\displaystyle 9\sqrt{6}\)

\(\displaystyle 3\sqrt{6}+3\)

\(\displaystyle 27\sqrt{2}\)

\(\displaystyle 3\sqrt{6}+9\)

\(\displaystyle 9\sqrt{2}+3\)

Correct answer:

\(\displaystyle 3\sqrt{6}+3\)

Explanation:

\(\displaystyle \sqrt{3}(3\sqrt{2}+\sqrt{3})\)

Order of operations, first distributing the \(\displaystyle \sqrt{3}\) to all terms inside the parentheses. 

\(\displaystyle (1\cdot 3\sqrt{3}\cdot \sqrt{2})+(\sqrt{3}\cdot\sqrt{3})\)

\(\displaystyle 3\sqrt{3\cdot 2}+\sqrt{3\cdot 3}\rightarrow3\sqrt{6}+\sqrt{9}\rightarrow3\sqrt{6}+3\)

The final answer is \(\displaystyle 3\sqrt{6}+3\).

Example Question #1 : How To Multiply Square Roots

The square root(s) of 36 is/are ________.

Possible Answers:

-6

6

6, -6, and 0

6 and -6

None of these answers are correct.

Correct answer:

6 and -6

Explanation:

To square a number is to multiply that number by itself. Because 6 x 6 = 36 AND -6 x -6 = 36, both 6 and -6 are square roots of 36.

Example Question #3 : How To Multiply Square Roots

Simplify:

\(\displaystyle \sqrt{4}*\sqrt{12}*\sqrt{5}\)

Possible Answers:

\(\displaystyle \sqrt{21}\)

\(\displaystyle 5\sqrt{12}\)

\(\displaystyle 2\sqrt{60}\)

\(\displaystyle 4\sqrt{15}\)

\(\displaystyle \sqrt{15}\)

Correct answer:

\(\displaystyle 4\sqrt{15}\)

Explanation:

Multiplication of square roots is easy! You just have to multiply their contents by each other. Just don't forget to put the result "under" a square root! Therefore:

\(\displaystyle \sqrt{4}*\sqrt{12}*\sqrt{5}\)

becomes

\(\displaystyle \sqrt{4*12*5}=\sqrt{240}\)

Now, you need to simplify this:

\(\displaystyle \sqrt{240} = \sqrt{2*2*2*2*3*5}\)

You can "pull out" two \(\displaystyle 2\)s.  (Note, that it would be even easier to do this problem if you factor immediately instead of finding out that \(\displaystyle 4*12*5 = 240\).)

After pulling out the \(\displaystyle 2\)s, you get:

\(\displaystyle \sqrt{2*2*2*2*3*5} = 2*2(\sqrt{3*5}) = 4\sqrt{15}\)

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