Simple algebra simplification

Question

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

Answer 1

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}

Answer 2

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

Answer 3

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}

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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.