How to simplify square roots

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PSAT Math › How to simplify square roots

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Simplify. Assume all integers are positive real numbers.

$\sqrt[3]{24x^{16}y^{12}}$

$2x^{5}y^{4}\sqrt[3]{3x}$

CORRECT

$8x^{16}y^{12}$

$2x^{8}y^{4}\sqrt[3]{3x^{2}}$

$2x^{2}y^{3}\sqrt[3]{3xy}$

$2x^{5}y^{4}$

Explanation

$\sqrt[3]{24x^{16}y^{12}}$

Index of means the cube root of Radican $24x^{16}y^{12}$

Find a perfect cube in $\rightarrow$ $\sqrt[3]{8\cdot 3}$

Simplify the perfect cube, giving you $2\sqrt[3]{3}$ .

Take your exponents on both variables and determine the number of times our index will evenly go into both.

$x\rightarrow \frac{16}{3}= 5^{\frac{1}{3}} \rightarrow x^{5}\sqrt[3]{x}$

$y\rightarrow \frac{12}{3}= 4 \rightarrow y^{4}$

The final answer would be

$2x^{5}y^{4}\sqrt[3]{3x}$

Simplify:

$4\sqrt{27}+16\sqrt{75}+3\sqrt{12}$

$98\sqrt{3}$

CORRECT

$80\sqrt{3}$

$16\sqrt{3}$

$92\sqrt{3}$

$96\sqrt{3}$

Explanation

4√27 + 16√75 +3√12 =

4*(√3)*(√9) + 16*(√3)*(√25) +3*(√3)*(√4) =

4*(√3)*(3) + 16*(√3)*(5) + 3*(√3)*(2) =

12√3 + 80√3 +6√3= 98√3

Simplify

9 ÷ √3

3√3

CORRECT

not possible

none of these

Explanation

in order to simplify a square root on the bottom, multiply top and bottom by the root

Asatsimplifysquare_root

Simplify:

√192

8√2

8√3

CORRECT

4√3

4√2

None of these

Explanation

√192 = √2 X √96

√96 = √2 X √48

√48 = √4 X√12

√12 = √4 X √3

√192 = √(2X2X4X4) X √3

= √4X√4X√4 X √3

= 8√3

What is the simplest way to express $\sqrt{3888}$ ?

$36\sqrt{3}$

CORRECT

$12\sqrt{27}$

$2\sqrt{972}$

$2304\sqrt{2}$

$144\sqrt{27}$

Explanation

First we will list the factors of 3888:

$3888=3\times1296=3\times\3\times432=3^2\times12\times36=3^2\times12\times12\times3=3^2\times12^2\times3$

$\sqrt{3888}=\sqrt{3^{2}\times 12^{2}\times 3}=\sqrt{3^{2}}\times\sqrt{12^{2}} \times \sqrt{3}=3\times 12\times \sqrt{3}=36\sqrt{3}$

Simplify: $\sqrt{72}$

$6\sqrt{2}$

CORRECT

$\sqrt{12}$

$2\sqrt{6}$

$8\sqrt{3}$

$7\sqrt{2}$

Explanation

To simplify a square root, you can break the number down into its prime factors using a factor tree. The prime factors of 72 are $2^{3}\times3^{2}$ . Let's take each piece separately.

The square root of $2^{3}$ can be simplified to be $\sqrt{2^{2}}\times \sqrt{2^{1}}$ which is the same as $2\sqrt{2}$ .

The square root of $3^{2}$ is .

When you multiply together your answers, $2\sqrt{2}\times3=6\sqrt{2}$

Simplify square roots. Assume all integers are positive real numbers.

Simplify as much as possible. List all possible answers.

1a. $\sqrt{12}$

1b. $\sqrt{48}$

1c. $\sqrt[3]{24}$

$2\sqrt{3}$ and $4\sqrt{3}$ and $2\sqrt[3]{3}$

CORRECT

$2\sqrt{3}$ and $4\sqrt{3}$ and $8\sqrt{3}$

$4\sqrt{3}$ and $\sqrt{3}$ and $2\sqrt[3]{3}$

$2\sqrt{3}$ and $4\sqrt{3}$

Explanation

When simplifying radicans (integers under the radical symbol), we first want to look for a perfect square. For example, $\sqrt{12}$ is not a perfect square. You look to find factors of to see if there is a perfect square factor in , which there is.

1a. $\sqrt{12}=\sqrt{4\cdot 3}=\sqrt{4}\cdot\sqrt{3}= 2\sqrt{3}$

Do the same thing for $\sqrt{48}$ .

1b. $\sqrt{48}=\sqrt{16\cdot 3}=\sqrt{16}\cdot\sqrt{3}=4\sqrt{3}$

1c.Follow the same procedure except now you are looking for perfect cubes.

$\sqrt[3]{24}=\sqrt[3]{8\cdot 3}=\sqrt[3]{8}\cdot\sqrt[3]{3}=2\sqrt[3]{3}$

Simplify. Assume all variables are positive real numbers.

$\sqrt{16x^{5}{y^{2}}}$

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

CORRECT

$4x^{5}y$

$4x^{5}y^{2}$

$4x^{2}y^{2}$

$4x^{2}y$

Explanation

$\sqrt{16x^{5}{y^{2}}}$

The index coefficent in $\sqrt[n]{b}$ is represented by . When no index is present, assume it is equal to 2. under the radical is known as the radican, the number you are taking a root of.