Subtracting two square roots How should one subtract the following square roots? I am aware of the fact that you can only add or subtract two square roots if their 'radical part' is the same.
$\displaystyle y^* = \sqrt{\frac{2KD}{h}} \cdot \sqrt{\frac{s+h}{s}}$
$\displaystyle y_s^* = \sqrt{\frac{2KD}{h}} \cdot \sqrt{\frac{s}{s+h}}$
$\displaystyle y^* - y_s^* = \ ???$
The answer should be $\displaystyle \sqrt{\frac{2KD}{s}} \cdot  \sqrt{\frac{h}{s+h}}$
 A: Let us start by writing
$$A = \frac{2KD}{h}$$
You want to simplify
$$\sqrt{A}\sqrt{\frac{s+h}{s}} - \sqrt{A}\sqrt{\frac{s}{s+h}} = \\
\sqrt{A}\left(\sqrt{\frac{s+h}{s}} - \sqrt{\frac{s}{s+h}} \right)
$$
Now let us rewrite $\sqrt{\frac{s}{s+h}}$:
$$\sqrt{\frac{s}{s+h}} = \sqrt{\frac{1}{\frac{s+h}{s}}} = \frac{1}{\sqrt{\frac{s+h}{s}}}$$
Put that back and reduce to the same denominator:
$$\sqrt{A}\left(\sqrt{\frac{s+h}{s}} - \sqrt{\frac{s}{s+h}} \right) = \\
\sqrt{A}\left(\sqrt{\frac{s+h}{s}} - \frac{1}{\sqrt{\frac{s+h}{s}}}\right) = \\
\sqrt{A}\left(\frac{\frac{s+h}{s}}{\sqrt{\frac{s+h}{s}}} - \frac{\frac{s}{s}}{\sqrt{\frac{s+h}{s}}}\right) = \\
\sqrt{A}\left( \frac{\frac{h}{s}}{\sqrt{\frac{s+h}{s}}} \right)
$$
Can you take it from here? You should put the $\frac{h}{s}$ inside the square root, write $A$ in its original form and manipulate one $h$ and one $s$ from the right factor to the $A$ factor.
A: Well, just..do the the Math.
We have : $$\sqrt\frac{2KD}{h}\sqrt\frac{s+h}{s}-\sqrt\frac{2KD}{h}\sqrt\frac{s}{s+h}=\\\sqrt\frac{2KD}{h}\Big(\sqrt\frac{s+h}{s}-\sqrt\frac{s}{s+h}\Big)=\\\sqrt\frac{2KD}{h}\Big(\frac{s+h-s}{\sqrt{s(s+h)}}\Big)=\\\sqrt\frac{2KD}{h}\Big(\frac{h}{\sqrt{s(s+h)}}\Big)=\\\sqrt{2KD}\frac{h}{\sqrt{h}}\frac1{\sqrt{s(s+h)}}=\sqrt\frac{2KD}{s}\sqrt\frac{h}{s+h}\\$$
since $$\frac{h}{\sqrt{h}}=\frac{h\sqrt{h}}{(\sqrt{h})^2}=\sqrt{h}$$
