GED Math - Square Roots and Radicals

Simplify $\sqrt{45}$ .

$3\sqrt{5}$

$9\sqrt{5}$

$5\sqrt{3}$

Explanation

When simplifying a square root, you must break up what's inside the square root into its simplest factors. For example would break up to $2\cdot5$ . Once you do that, you look for pairs of numbers. For each pair, you pull out the common number to the outside of the square root and leave whatever is left over inside the square root. So, for $\sqrt{45}$ , you would break that up to $9\cdot5$ and then $3\cdot3\cdot5$ . As you can see, there is a pair of , so you pull out a and leave whatever is left inside. Since the is left over you would leave that inside the square root. So, you have a outside the square root and a inside the square root, which gives you $3\sqrt{5}$ .

Simplify:
$\small \sqrt{192}$

$\small \small 8\sqrt3$

$\small 64\sqrt3$

$\small 4\sqrt{12}$

$\small 16\sqrt{12}$

Explanation

$\small \sqrt{192}=\sqrt{64}\times \sqrt3=8\times \sqrt3=8\sqrt3$

An alternate solution is:

$\small \sqrt{192}=\sqrt{16}\times \sqrt{12}=4\times \sqrt{4} \times \sqrt3=4\times2\times\sqrt3=8\sqrt3$

Simplify:

$\small \sqrt{x^4}$

$\small x^2$

$\small x^3$

$\small x$

$\small \sqrt{x^4}$

Explanation

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

$m = \sqrt{7}$

$n = \sqrt{9}$

$p = \sqrt{11}$

$q= \sqrt{13}$

How many of , , , and are irrational numbers?

Explanation

The square root of an integer is a rational number if and only if the radicand - the number under the symbol - is itself a perfect square. Of the integers under the four square roots given, only 9 is a perfect square, being equal to $3 \times 3 = 3^{2 }$ . The other three numbers are therefore irrational.

Rationalize: $\frac{6}{5\sqrt6}$

$\frac{\sqrt6}{5}$

$\frac{3\sqrt6}{5}$

$\frac{\sqrt3}{5}$

$\frac{\sqrt6}{30}$

$\frac{2\sqrt6}{5}$

Explanation

Multiply the radical on the top and bottom of the fraction.

$\frac{6}{5\sqrt6} \cdot \frac{\sqrt6}{\sqrt6} = \frac{6\sqrt6}{5(6)}$

Reduce the fraction.

The answer is: $\frac{\sqrt6}{5}$

Solve for $\small z$ : $\small \small \small \sqrt{z}=x^4$

$\small \small z=x^8$

$\small \small z=\sqrt{x}$

$\small \small z=x^2$

$\small z=\sqrt{x^4}$

$\small z=2x$

Explanation

In order to solve for $\small z$ , we need to move all of the variables on its side of the equation over to the other side of the equation.

We can see that our $\small z$ is hiding in a square root, so in order to get the $\small z$ out we will need to square the whole equation.

$\small \small \small \sqrt{z}=x^4$

$\small (\sqrt{z})^2=(x^4)^2$

Because we're multiply a power of $\small 2$ with the power of $\small 4$ , the two can multiply together to create a power of $\small 8$

$\small z=x^8$

We can't do anything else to this equation as there are no like variables.

Our answer is $\small z=x^8$

Factor: $6\sqrt{40}$

$12\sqrt{10}$

$\textup{The expression is already factored.}$

$2\sqrt{30}$

$24\sqrt{10}$

$40\sqrt{3}$

Explanation

Rewrite the root 40 in factors of perfect squares.

$6\sqrt{40} = 6 \cdot \sqrt{4}\cdot \sqrt{10} = 6 \cdot 2\cdot \sqrt{10}$

The answer is: $12\sqrt{10}$

Rationalize: $\frac{2}{\sqrt{12}}$

$\frac{\sqrt3}{3}$

$\frac{2\sqrt3}{3}$

$\frac{2\sqrt3}{5}$

$6\sqrt3$

$6\sqrt2$

Explanation

Factor the denominator by factors of perfect squares.

$\sqrt{12} = \sqrt4 \cdot \sqrt3 = 2\sqrt{3}$

Replace the term.

$\frac{2}{\sqrt{12}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt3}$

Multiply by root three on the top and bottom.

$\frac{1}{\sqrt3} \cdot \frac{\sqrt3}{\sqrt3} = \frac{\sqrt3}{3}$

The answer is: $\frac{\sqrt3}{3}$

Simplify the following expression:

$7\sqrt{3}+\sqrt{75}+\sqrt{243}$

$21\sqrt{3}$

$14\sqrt{3}+2\sqrt{2}$

$12\sqrt{3}+12$

$\sqrt{3}+5\sqrt{6}$

Explanation

Start by analyzing each given term.

$7\sqrt{3}$ cannot be reduced any further, so leave it alone.

Notice that $\sqrt{75}$ can be rewritten as $\sqrt{25(3)}=5\sqrt3$

Next, notice that $\sqrt{243}$ can be rewritten as $\sqrt{(81)(3)}=9\sqrt{3}$

Now, rewrite the original equation:

$7\sqrt{3}+\sqrt{75}+\sqrt{243}=7\sqrt{3}+5\sqrt{3}+9\sqrt{3}=21\sqrt{3}$

Simplify the following expression:

$7\sqrt{5}+\sqrt{20}+\sqrt{500}$

$19\sqrt5$

$7\sqrt{5}+4$

$12\sqrt{5}+10\sqrt{2}$

$21\sqrt{3}$

Explanation

Start by putting each term in terms of $\sqrt{5}$ .

$7\sqrt{5}$ is already in terms of $\sqrt{5}$ so leave it alone.

Notice that $\sqrt{20}$ can be rewritten as $\sqrt{4\times 5}$ . Thus $\sqrt{20}=2\sqrt{5}$ .

Next, notice that $\sqrt{500}$ can be rewritten as $\sqrt{100\times 5}$ . Thus $\sqrt{500}=10\sqrt{5}$ .

Now, add these terms together.

$7\sqrt{5}+2\sqrt{5}+10\sqrt{5}=19\sqrt{5}$

Square Roots and Radicals

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