Square Roots and Radicals - GED Math

Card 0 of 20

Question

Simplify:
$\small \sqrt{192}$

Answer

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

An alternate solution is:

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

Compare your answer with the correct one above

Question

Simplify:

$\small \sqrt{x^4}$

Answer

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

Compare your answer with the correct one above

Question

Simplify:

$\small \sqrt{98}$

Answer

$\small \sqrt{98}=\sqrt{49}\times \sqrt{2}=7\times \sqrt2=7\sqrt2$

Compare your answer with the correct one above

Question

Simplify:

$\small \sqrt{72}=$

Answer

$\small \sqrt{72}=\sqrt{36}\times \sqrt2=6\times \sqrt2=6\sqrt2$

Compare your answer with the correct one above

Question

Simplify:

$\sqrt{147}$

Answer

$\sqrt{147}=\sqrt{49}\times \sqrt3=7\times \sqrt3=7\sqrt3$

Compare your answer with the correct one above

Question

Refer to the above number line. What point most likely represents the square root of 280?

Do not use a calculator.

Answer

$16^{2} = 256 < 280 < 289 < 17^{2}$

Therefore, $16 < \sqrt{280} < 17$ ,

making the correct choice .

Compare your answer with the correct one above

Question

Simplify:

$\sqrt{42}$

Do not use a calculator.

Answer

The prime factorization of 42 is

$42 = 2 \times 3 \times 7$ .

Since 42 is the product of distinct primes, it has no perfect square factors, and, therefore, its square root cannot be simplified further. It is already in simplifed form.

Compare your answer with the correct one above

Question

Simplify:

$\sqrt{48}$

Do not use a calculator.

Answer

The prime factorization of 48 is

$48 = 2 \times 2 \times 2 \times 2 \times 3$ .

Rewrite, and use the product of radicals property to simplify:

$\sqrt{48}$

$= \sqrt{ 2 \times 2 \times 2 \times 2 \times 3}$

$= \sqrt{ 2 \times 2 } \times \sqrt{ 2 \times 2 } \times \sqrt{ 3}$

$= 2 \times 2 \times \sqrt{ 3}$

$=4 \sqrt{ 3}$

Compare your answer with the correct one above

Question

Factor: $2\sqrt{120}$

Answer

In order to factor the radical, we will need to rewrite 120 as multiples of perfect squares.

$2\sqrt{120} = 2\sqrt{30} \cdot \sqrt{4}$

Reduce the known term

$2\sqrt{30} \cdot \sqrt{4} =2\sqrt{30} \cdot 2 = 4\sqrt{30}$

The value of $\sqrt{30}$ cannot be factored any further.

The answer is: $4\sqrt{30}$

Compare your answer with the correct one above

Question

Factor: $6\sqrt{40}$

Answer

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: $12\sqrt{10}$

Compare your answer with the correct one above

Question

Factor: $3\sqrt{88}$

Answer

In order to factor this, we will need to rewrite root 88 by factoring using values of perfect squares.

$3\sqrt{88} = 3\times\sqrt{4}\times \sqrt{22} = 3\times 2 \times \sqrt{22}$

The answer is: $6\sqrt{22}$

Compare your answer with the correct one above

Question

Rationalize: $\frac{3}{2\sqrt2}$

Answer

In order to rationalize the radical, we will need to multiply both the top and bottom by square root two.

$\frac{3}{2\sqrt2}\cdot\frac{\sqrt2}{\sqrt2} = \frac{3\sqrt2}{2\cdot 2}$

The answer is: $\frac{3\sqrt{2}}{4}$

Compare your answer with the correct one above

Question

Rationalize: $\frac{2}{\sqrt5}$

Answer

In order to eliminate the radical on the denominator, we will need to multiply root five on the top and bottom of the fraction.

$\frac{2}{\sqrt5}\cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$

The answer is: $\frac{2\sqrt{5}}{5}$

Compare your answer with the correct one above

Question

Rationalize: $\frac{\sqrt3}{\sqrt{20}}$

Answer

Simplify the denominator by factoring using perfect squares.

$\sqrt{20} = \sqrt4 \cdot \sqrt5 = 2\sqrt{5}$

Rewrite the fraction.

$\frac{\sqrt3}{\sqrt{20}}=\frac{\sqrt3}{2\sqrt{5}}$

Multiply the top and bottom by root 5.

$\frac{\sqrt3}{2\sqrt{5}} \cdot \frac{\sqrt5}{\sqrt5} = \frac{\sqrt{15}}{10}$

The answer is: $\frac{\sqrt{15}}{10}$

Compare your answer with the correct one above

Question

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

Answer

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: $\frac{\sqrt3}{3}$

Compare your answer with the correct one above

Question

Rationalize: $\frac{7\sqrt2}{\sqrt{14}}$

Answer

Multiply by root 14 on the top and bottom of the fraction.

$\frac{7\sqrt2}{\sqrt{14}}\cdot\frac{\sqrt{14}}{\sqrt{14}} = \frac{7\sqrt{28}}{14}$

The fraction becomes: $\frac{\sqrt{28}}{2}$

Simplify the radical with factors of perfect squares.

$\frac{\sqrt{28}}{2} = \frac{\sqrt{4}\cdot \sqrt{7}}{2} = \frac{2\cdot \sqrt{7}}{2}$

Reduce the fraction.

The answer is: $\sqrt7$

Compare your answer with the correct one above

Question

Rationalize the radical: $\frac{9}{\sqrt3}$

Answer

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

$\frac{9}{\sqrt3} \cdot \frac{\sqrt3}{\sqrt3} = \frac{9\sqrt3}{3}$

Simplify the fraction.

The answer is: $3\sqrt3$

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

Rationalize: $\frac{5\sqrt2}{\sqrt3}$

Answer

Multiply the denominator on the top and bottom.

$\frac{5\sqrt2}{\sqrt3}\cdot \frac{\sqrt3}{\sqrt3} = \frac{5\sqrt6}{3}$

The answer is: $\frac{5\sqrt6}{3}$

Compare your answer with the correct one above

Question

Simplify the following expression:

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

Answer

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:

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

Compare your answer with the correct one above

Tap the card to reveal the answer