ACT Math : How to find the common factors of squares

Study concepts, example questions & explanations for ACT Math

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Example Questions

Example Question #1 : Simplifying Square Roots

Simplify:

\displaystyle \sqrt[2]{24,300}

Possible Answers:

\displaystyle 90\sqrt[2]{3}

\displaystyle 9\sqrt[2]{300}

\displaystyle 10\sqrt[2]{243}

\displaystyle 90\sqrt[2]{270}

\displaystyle 900\sqrt[2]{3}

Correct answer:

\displaystyle 90\sqrt[2]{3}

Explanation:

\displaystyle \sqrt[2]{24,300}

When simplifying the square root of a number that may not have a whole number root, it's helpful to approach the problem by finding common factors of the number inside the radicand. In this case, the number is 24,300.

What are the factors of 24,300?

24,300 can be factored into:

\displaystyle 24,300: 243\cdot 100

When there are factors that appear twice, they may be pulled out of the radicand. For instance, 100 is a multiple of 24,300. When 100 is further factored, it is \displaystyle 10^{2} (or 10x10). However, 100 wouldn't be pulled out of the radicand, but the square root of 100 because the square root of 24,300 is being taken. The 100 is part of the24,300. This means that the problem would be rewritten as: \displaystyle 10\sqrt[2]{243} 

But 243 can also be factored: \displaystyle 24,300: (3\cdot 81)\cdot100
\displaystyle 24,300: {\color{Orange} 9}\cdot{\color{Orange} 9}\cdot {\color{Cyan} 10}\cdot {\color{cyan} 10}\cdot 3
Following the same principle as for the 100, the problem would become
\displaystyle 9\cdot 10\sqrt[2]{3} because there is only one factor of 3 left in the radicand. If there were another, the radicand would be lost and it would be 9*10*3. 
9 and 10 may be multiplied together, yielding the final simplified answer of 
\displaystyle 90\sqrt[2]{3}

Example Question #1 : Arithmetic

\displaystyle \sqrt{180} + \sqrt{125} = ?

Possible Answers:

\displaystyle 25.0

\displaystyle 11\sqrt{5}

\displaystyle 11\sqrt{10}

\displaystyle 17.5

\displaystyle \sqrt{305}

Correct answer:

\displaystyle 11\sqrt{5}

Explanation:

To solve the equation \displaystyle \sqrt{180} + \sqrt{125} = ?, we can first factor the numbers under the square roots.

\displaystyle \sqrt{2\cdot 2\cdot 3\cdot 3\cdot 5} + \sqrt{5\cdot 5\cdot 5} = ?

When a factor appears twice, we can take it out of the square root.

\displaystyle 2\cdot 3\sqrt{5} + 5\sqrt{5} = ?

\displaystyle 6\sqrt{5} + 5\sqrt{5} = ?

Now the numbers can be added directly because the expressions under the square roots match.

\displaystyle 6\sqrt{5} + 5\sqrt{5} = 11\sqrt{5}

 

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