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
[tex]\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}\\\\[/tex]
In the third step, I multiplied top and bottom by [tex]\sqrt{7}[/tex] so that the denominator can get rid of the square root.
Answer: [tex]\frac{\sqrt{21\text{x}}}{7}\\\\[/tex]
======================================================
Problem 7
We'll follow the same idea for this problem.
[tex]\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}\\\\[/tex]
Where m is nonnegative.
In the second to last step, I split up the root so I could use the rule [tex]\sqrt{x^2} = x \text{ when } x \ge 0[/tex]
Answer: [tex]\frac{m\sqrt{10}}{2}\\\\[/tex]
======================================================
Problem 8
[tex]\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}\\\\[/tex]
where x is nonnegative.
Answer: [tex]\frac{2x\sqrt{15\text{x}}}{5}\\\\[/tex]
