Question

just please help 6 through 8 thank you please help me its a homework but i really dont get it that much please help me​

Answer 1

Problem 6

It's not clear what the instructions are, but I'm assuming your teacher wants you to rationalize the denominator.

If so, then we could have these steps

$\sqrt{\frac{3\text{x}}{7}}\\\\\frac{\sqrt{3\text{x}}}{\sqrt{7}}\\\\\frac{\sqrt{3\text{x}}*\sqrt{7}}{\sqrt{7}*\sqrt{7}}\\\\\frac{\sqrt{3\text{x}*7}}{(\sqrt{7})^2}\\\\\frac{\sqrt{21\text{x}}}{7}$

In the third step, I multiplied top and bottom by $\sqrt{7}$ so that the denominator can get rid of the square root.

Answer: $\frac{\sqrt{21\text{x}}}{7}$

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Problem 7

We'll follow the same idea for this problem.

$\sqrt{\frac{5m^2}{2}}\\\\\frac{\sqrt{5m^2}}{\sqrt{2}}\\\\\frac{\sqrt{5m^2}*\sqrt{2}}{\sqrt{2}*\sqrt{2}}\\\\\frac{\sqrt{5m^2*2}}{(\sqrt{2})^2}\\\\\frac{\sqrt{10m^2}}{2}\\\\\frac{\sqrt{10}*\sqrt{m^2}}{2}\\\\\frac{m\sqrt{10}}{2}$

Where m is nonnegative.

In the second to last step, I split up the root so I could use the rule $\sqrt{x^2} = x \text{ when } x \ge 0$

Answer:  $\frac{m\sqrt{10}}{2}$

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Problem 8

$\sqrt{\frac{12\text{x}^3}{5}}\\\\\frac{\sqrt{12\text{x}^3}}{\sqrt{5}}\\\\\frac{\sqrt{12\text{x}^3}*\sqrt{5}}{\sqrt{5}*\sqrt{5}}\\\\\frac{\sqrt{12\text{x}^3*5}}{(\sqrt{5})^2}\\\\\frac{\sqrt{60\text{x}^3}}{5}\\\\\frac{\sqrt{4\text{x}^2*15\text{x}}}{5}\\\\\frac{\sqrt{4\text{x}^2}*\sqrt{15\text{x}}}{5}\\\\\frac{\sqrt{(2\text{x})^2}*\sqrt{15\text{x}}}{5}\\\\\frac{2x\sqrt{15\text{x}}}{5}$

where x is nonnegative.

Answer: $\frac{2x\sqrt{15\text{x}}}{5}$