Question

Solve the following equations:
$\sqrt{x^2+x}+\frac{\sqrt{x-1}}{\sqrt{x^3-x}}=\frac{5}{2}$.

Answer 1

$\sqrt{x^2+x}+\frac{\sqrt{x-1}}{\sqrt{x^3-x}}=\frac{5}{2}\sqrt{x^2+x}+\frac{\sqrt{x-1}}{\sqrt{x(x^2-1)}}=\frac{5}{2}\sqrt{x^2+x}+\frac{\sqrt{x-1}}{\sqrt{x(x-1)(x+1)}}=\frac{5}{2}\sqrt{x^2+x}+\frac{1}{\sqrt{x^2+x}}=\frac{5}{2}$

Put $\sqrt{x^2+x}=t$

$t+\frac{1}{t}=\frac{5}{2}\frac{t^2+1}{t}=\frac{5}{2}2t^2-5t+2=02t^2-4t-t+2=02t(t-2)-1(t-2)=0(2t-1)(t-2)=0t=\frac{1}{2},2$

$x^2+x=t^2{x}^2+x={\left(\frac{1}{2}\right)}^{2}4x^2+4x-1=\mathrm{0........}\left(i\right)$

using quadratic formula

$x=\frac{-4\pm \sqrt{16-4\left(4\right)(-1)}}{2\left(4\right)}=\frac{-4\pm \sqrt{32}}{8}x=\frac{-4\pm 4\sqrt{2}}{8}=\frac{-1\pm \sqrt{2}}{2}$

Also $x^2+x=2^2$

$x^2+x-4=\mathrm{0......}\left(ii\right)$
using quadratic formula

$x=\frac{-1\pm \sqrt{1-4(-4)}}{2}=\frac{-1\pm \sqrt{17}}{2}$

So the values of $x$ are $\frac{-1\pm \sqrt{2}}{2}$ and $\frac{-1\pm \sqrt{17}}{2}$