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I have

$ \frac{-8rS^{2} \pm \sqrt{64r^{2}S^{4} + 4}}{2}$

Which simplified to

$ -4rS^{2} \pm \sqrt{16r^{2}S^{4} +1}$

But I can’t seem to get this

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    $\begingroup$ $$\pm \frac{1}{2} = \sqrt{\frac{1}{4}}$$ $\endgroup$ – Kevin Apr 24 '18 at 11:17
  • $\begingroup$ hint: $\frac{\sqrt{64}}{2} = \frac{\sqrt{64}}{\sqrt{4}} = \sqrt{\frac{64}{4}} \dots $ $\endgroup$ – Matti P. Apr 24 '18 at 11:17
  • $\begingroup$ I don’t get it, can you help from here $\endgroup$ – italy Apr 24 '18 at 11:38
  • $\begingroup$ I don't see what I am doing wrong, $ \frac{\sqrt{64 r^{2}S^{4}+4}}{2}$ = $ \frac{\sqrt{64}\sqrt{r^2S^4}}{2}$ + \frac{\sqrt{4}}{2}$ $\endgroup$ – italy Apr 24 '18 at 11:45
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Let $$D = \frac{-8rS^{2} \pm \sqrt{64r^{2}S^{4} + 4}}{2}$$

Now \begin{align} D &= \frac{-8rS^{2} \pm \sqrt{64r^{2}S^{4} + 4}}{2}\\ &=\frac{-8rS^{2}}{2} \pm \frac{\sqrt{64r^{2}S^{4} + 4}}{2} \\ &= -4rS^2 \pm \frac{\sqrt{64r^{2}S^{4} + 4}}{\sqrt{4}} \\ &= -4rS^2 \pm \sqrt{\frac{64r^{2}S^{4} + 4}{4}}\\ &=-4rS^2 \pm \sqrt{16r^{2}S^{4} + 1} \end{align}

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  • $\begingroup$ Thank you, but can you also explain why its not acceptable to just do this $ \frac{\sqrt{64} \sqrt{r^{2}S^{4}}{2} $ + $ \frac{\sqrt{4}}{2}$ =? $ 4\sqrt{r^2}{S^{4} + 1 $ $\endgroup$ – italy Apr 24 '18 at 11:56
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    $\begingroup$ @italy you cannot split a square root like this. If you could, you'd get some very weird results - for example $$\sqrt 2=\sqrt{1+1}=\sqrt1+\sqrt1=1+1=2$$When you spread an operator over a sum like this, it is called the distributive property. It does not hold for the square root operator. $\endgroup$ – John Doe Apr 24 '18 at 12:34
  • $\begingroup$ @italy John Doe's comment above this is the reason why. Do you see now? $\endgroup$ – Kevin Apr 24 '18 at 15:24
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Note that $$ 64r^2S^4+4=4(16r^2S^4+1) $$ so $$ \sqrt{64r^2S^4+4}=\sqrt{4(16r^2S^4+1)}=2\sqrt{16r^2S^4+1} $$ Finally, \begin{align} \frac{-8rS^2\pm\sqrt{64r^2S^4+4}}{2} &=\frac{-8rS^2\pm2\sqrt{16r^2S^4+1}}{2}\\[6px] &=\frac{2(-4rS^2\pm\sqrt{16r^2S^4+1}\,)}{2}\\[6px] &=-4rS^2\pm\sqrt{16r^2S^4+1} \end{align}


Your original equation apparently was $$ x^2+8rS^2x-1=0 $$ Whenever you have a factor $2$ that can be extracted from the coefficient of the degree $1$ term, the simplification above can be performed: if the equation is $ax^2+2\beta x+c=0$, the quadratic formula yields \begin{align} \frac{-2\beta\pm\sqrt{4\beta^2-4ac}}{2a} &=\frac{-2\beta\pm\sqrt{4(\beta^2-ac)}\,}{2a}\\[6px] &=\frac{-2\beta\pm2\sqrt{\beta^2-ac}}{2a}\\[6px] &=\frac{-\beta\pm\sqrt{\beta^2-ac}}{a} \end{align} In your case $a=1$, $\beta=-4rS^2$ and $c=-1$.

It's not necessary to remember one more formula, just to recall that the simplification can be done.

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We have $\frac{-8rS^2}{2}=-4rS^2$ and $\frac{\sqrt{4X}}{2}=\sqrt{X}$.

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  • $\begingroup$ This answer could be helped by doing that last simplification over an extra step ($2=\sqrt 4$ for example). That seems to be where OP's confusion lies. $\endgroup$ – John Doe Apr 24 '18 at 11:19
  • $\begingroup$ @JohnDoe: seeing his comment, the OP believes in the distributivity of the square root over addition ! $\endgroup$ – Yves Daoust Apr 24 '18 at 11:58
  • $\begingroup$ @YvesDaoust Ohh right I see, well thats a completely different issue! $\endgroup$ – John Doe Apr 24 '18 at 12:31

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