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 $\displaystyle \sqrt{\frac{2KD}{s}} \cdot \sqrt{\frac{h}{s+h}}$

  • $\begingroup$ Are all your variables, in particular $h$ and $s$, assumed to be positive? $\endgroup$ – Barry Cipra Jan 10 '17 at 21:34
  • $\begingroup$ Yeah, all variables are > 0. $\endgroup$ – Anna Jan 10 '17 at 21:42

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 $$\frac{h}{\sqrt{h}}=\frac{h\sqrt{h}}{(\sqrt{h})^2}=\sqrt{h}$$

  • $\begingroup$ What do you do in the third step? Subtracting of the two roots is not possible there right (since the 'radical parts' are not equal)? $\endgroup$ – Anna Jan 10 '17 at 21:21
  • $\begingroup$ @Anna In more detail $\frac{\sqrt{s+h}}{\sqrt{s}}-\frac{\sqrt{s}}{\sqrt{s+h}}=\frac{(\sqrt{s+h})^2}{\sqrt{s}\sqrt{s+h}}-\frac{(\sqrt{s})^2}{\sqrt{s}\sqrt{s+h}}$ I simply make the denominators equal.. $\endgroup$ – MathematicianByMistake Jan 10 '17 at 21:26
  • 1
    $\begingroup$ Perfectly clear. $\endgroup$ – Anna Jan 10 '17 at 21:57
  • $\begingroup$ @Anna Glad to help! $\endgroup$ – MathematicianByMistake Jan 10 '17 at 22:07

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.

  • $\begingroup$ Why is your third step allowed? In general, you can't just take the reciprocal (1/...) of just one part of the expression? $\endgroup$ – Anna Jan 10 '17 at 21:20
  • $\begingroup$ @Anna notice I took the reciprocal 2 times. Just like you can add and subtract some number, $k = k + x - x$ or divide and multiply by the same number, $k = k\cdot x\cdot \frac1x$, you can take the reciprocal 2 times: $k = \frac{1}{\frac{1}{k}}$. In my case the second reciprocal just flips the fraction. Can you take it from where I left it or should I finish it? $\endgroup$ – RGS Jan 10 '17 at 21:26
  • $\begingroup$ Thanks for the clarification (and the intuitive steps). I did manage to work it through. $\endgroup$ – Anna Jan 10 '17 at 21:58

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