# Simple algebra simplification

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

• $$\pm \frac{1}{2} = \sqrt{\frac{1}{4}}$$ – Kevin Apr 24 '18 at 11:17
• hint: $\frac{\sqrt{64}}{2} = \frac{\sqrt{64}}{\sqrt{4}} = \sqrt{\frac{64}{4}} \dots$ – Matti P. Apr 24 '18 at 11:17
• I don’t get it, can you help from here – italy Apr 24 '18 at 11:38
• 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}– italy Apr 24 '18 at 11:45 ## 3 Answers 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} • 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 – italy Apr 24 '18 at 11:56 • @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. – John Doe Apr 24 '18 at 12:34 • @italy John Doe's comment above this is the reason why. Do you see now? – Kevin Apr 24 '18 at 15:24 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 factor2$that can be extracted from the coefficient of the degree$1$term, the simplification above can be performed: if the equation is$ax^2+2\beta x+c=0, the quadratic formula yields \begin{align} \frac{-2\beta\pm\sqrt{4\beta^2-4ac}}{2a} &=\frac{-2\beta\pm\sqrt{4(\beta^2-ac)}\,}{2a}\\[6px] &=\frac{-2\beta\pm2\sqrt{\beta^2-ac}}{2a}\\[6px] &=\frac{-\beta\pm\sqrt{\beta^2-ac}}{a} \end{align} In your casea=1$,$\beta=-4rS^2$and$c=-1$. It's not necessary to remember one more formula, just to recall that the simplification can be done. We have$\frac{-8rS^2}{2}=-4rS^2$and$\frac{\sqrt{4X}}{2}=\sqrt{X}$. • 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. – John Doe Apr 24 '18 at 11:19
• @JohnDoe: seeing his comment, the OP believes in the distributivity of the square root over addition ! – Yves Daoust Apr 24 '18 at 11:58
• @YvesDaoust Ohh right I see, well thats a completely different issue! – John Doe Apr 24 '18 at 12:31