Simplify the top of the fraction.

$\displaystyle \frac{\sqrt{3}\times \sqrt{5}}{\sqrt{32}}=\frac{\sqrt{15}}{\sqrt{32}}$

We can simplify the denominator first.

$\displaystyle \sqrt{32} = \sqrt{16}\times \sqrt2 =4\sqrt2$

Rationalize the denominator by multiplying the top and bottom of the fraction by the radical.  Simplify the radicals.

$\displaystyle \frac{\sqrt{15}}{4\sqrt2}\cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{30}}{4 \times 2}= \frac{\sqrt{30}}{8}$

The answer is:  $\displaystyle \frac{\sqrt{30}}{8}$

Rationalize the denominator by multiplying both the top and bottom by the denominator.

$\displaystyle \frac{4\sqrt{8}}{5\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{24}}{5\times3} = \frac{4\sqrt{4}\times \sqrt{6}}{15}=\frac{4(2)\times \sqrt{6}}{15}$

The answer is:  $\displaystyle \frac{8\sqrt{6}}{15}$

The radical $\displaystyle \sqrt{\frac{1}{7}}$ can be rewritten as:

$\displaystyle \frac{\sqrt1}{\sqrt7} = \frac{1}{\sqrt7}$

Rationalize the denominator by multiplying both the top and bottom by the denominator.

$\displaystyle \frac{1}{\sqrt7} \cdot \frac{\sqrt7}{\sqrt7} =\frac{\sqrt7}{7}$

Rewrite the given question.

$\displaystyle \frac{7\sqrt7}{\sqrt{\frac{1}{7}}} = \frac{7\sqrt7}{\frac{\sqrt7}{7}}$

Simplify this complex fraction by rewriting it as a division sign.

$\displaystyle \frac{7\sqrt7}{\frac{\sqrt7}{7}} = 7\sqrt7 \div \frac{\sqrt7}{7}$

Change the division sign to multiplication and take the reciprocal of the second term.

$\displaystyle 7\sqrt7 \div \frac{\sqrt7}{7} = 7\sqrt7 \times \frac{7}{\sqrt7} =49$

The answer is:  $\displaystyle 49$

We can factor the square root twenty before rationalizing the value so that the final term would not be as difficult to factor.

Factor $\displaystyle \sqrt{20}$ using factors of perfect squares.

$\displaystyle \sqrt{20}= \sqrt{4}\times \sqrt5 = 2\sqrt5$

The expression becomes:  

$\displaystyle \frac{9}{\sqrt{20}} = \frac{9}{2\sqrt{5}}$

Instead of multiplying $\displaystyle \sqrt{20}$ on both the top and bottom, we will multiply $\displaystyle \sqrt5$ on the top and bottom.

$\displaystyle \frac{9}{2\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{9\sqrt{5}}{10}$

The answer is:  $\displaystyle \frac{9\sqrt{5}}{10}$

First, apply the Quotient of Radicals Property to split the radical into a numerator and a denominator:

$\displaystyle \sqrt{\frac{5}{8}}$

$\displaystyle = \frac{\sqrt{5}}{\sqrt{8}}$

Simplify the denominator by applying the Product of Radicals Property, as follows:

$\displaystyle = \frac{\sqrt{5}}{\sqrt{4} \cdot \sqrt{2}}$

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

Rationalize the denominator by multiplying both halves of the fraction by $\displaystyle \sqrt{2}$:

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

Apply the Product of Radicals property in the numerator:

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

$\displaystyle = \frac{\sqrt{10}}{4}$,

the correct response.

First, apply the Quotient of Radicals Property to split the radical into a numerator and a denominator:

$\displaystyle \sqrt{\frac{12}{7}}$

$\displaystyle = \frac{\sqrt{12}}{\sqrt{7}}$

Simplify the numerator by applying the Product of Radicals Property, as follows:

$\displaystyle \frac{\sqrt{4 \cdot 3 }}{\sqrt{7}}$

$\displaystyle = \frac{\sqrt{4} \cdot \sqrt{3}}{\sqrt{7}}$

$\displaystyle = \frac{2 \sqrt{3}}{\sqrt{7}}$

Rationalize the denominator by multiplying both halves of the fraction by $\displaystyle \sqrt{7}$:

$\displaystyle \frac{2 \sqrt{3} \cdot \sqrt{7} }{\sqrt{7} \cdot \sqrt{7}}$

Apply the Product of Radicals Property in the numerator:

$\displaystyle \frac{2 \sqrt{3 \cdot 7} }{7}$

$\displaystyle = \frac{2 \sqrt{21} }{7}$

One way to solve this equation is to substitute $\displaystyle u$ for $\displaystyle \sqrt{x}$ and, subsequently, $\displaystyle u^{2}$ for $\displaystyle x$:

$\displaystyle x + 3 \sqrt{x} = 28$

$\displaystyle u^{2} + 3 u = 28$

$\displaystyle u^{2} + 3 u - 28 = 0$

Solve the resulting quadratic equation by factoring the expression:

$\displaystyle (u+7)(u-4) = 0$

Set each linear binomial to sero and solve:

$\displaystyle u + 7 = 0$

$\displaystyle u = -7$

or 

$\displaystyle u - 4 = 0$

$\displaystyle u = 4$

Substitute back:

$\displaystyle u = -7$

$\displaystyle \sqrt{x}= -7$ - this is not possible.

$\displaystyle u = 4$

$\displaystyle \sqrt{x}= 4$

$\displaystyle x = 16$ - this is the only solution.

None of the responses state that $\displaystyle x = 16$ is the only solution.

We can simplify the fraction:

$\displaystyle \frac{10\sqrt{8}}{\sqrt{16}} = \frac{10\sqrt{4\times 2}}{4}=\frac{10(2\sqrt{2})}{4} =\frac{20\sqrt{2}}{4}=5\sqrt{2}$

Plugging this into the equation leaves us with:

$\displaystyle 5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}$ 

Note: Because they are like terms, we can add them.

In order to solve this equation, we need to know that

             $\displaystyle (\sqrt{x-6})^{2} = x-6$

How? Because of these two facts: 

1. $\displaystyle \sqrt{x-6} = (x-6)^{\frac{1}{2}}$
2. Power rule of exponents: when we raise a power to a power, we need to mulitply the exponents:

                                      $\displaystyle \frac{1}{2} \times2=1$

With this in mind, we can solve the equation:

$\displaystyle \sqrt{x-6} =2$

In order to eliminate the radical, we have to square it. What we do on one side, we must do on the other.

$\displaystyle (\sqrt{x-6})^{2} = 2^{2}$

$\displaystyle x-6=4$

$\displaystyle x=10$

In order to solve this equation, we need to know that 

 $\displaystyle \bigg(\sqrt{\frac{x}{2}}\bigg)^{2} =\frac{x}{2}$

Note: This is due to the power rule of exponents.

With this in mind, we can solve the equation:

$\displaystyle \sqrt{\frac{x}{2}} =5$

In order to get rid of the radical we square it. Remember what we do on one side, we must do on the other.

$\displaystyle \bigg(\sqrt{\frac{x}{2}}\bigg)^{2} =5^{2}$

$\displaystyle \frac{x}{2} =25$

$\displaystyle \frac{2}{1} \bigg(\frac{x}{2}\bigg)=25\bigg(\frac{2}{1}\bigg)$

$\displaystyle x=50$