Question

Solve the following equations:
$\sqrt{2x^2-9x+4}+3\sqrt{2x-1}=\sqrt{2x^2+21x-11}$.

Answer 1

$\sqrt{2x^2-9x+4}+3\sqrt{2x-1}=\sqrt{2x^2+21x-11}\sqrt{2x^2-9x+4}+3\sqrt{2x-1}-\sqrt{2x^2+21x-11}=0\sqrt{2x^2-8x-x+4}+3\sqrt{2x-1}-\sqrt{2x^2-x+22x-11}=0\sqrt{2x(x-4)-1(x-4)}+3\sqrt{2x-1}-\sqrt{x(2x-1)+11(2x-1)}=0\sqrt{(2x-1)(x-4)}+3\sqrt{2x-1}-\sqrt{(x+11)(2x-1)}=0\sqrt{2x-1}(\sqrt{x-4}+3-\sqrt{x+11)})=0$

$\Rightarrow \sqrt{2x-1}=0$ $\Rightarrow x=\frac{1}{2}$

and $\sqrt{x-4}+3-\sqrt{x+11}=0$

$\sqrt{x-4}+3-\sqrt{x+11}=0\sqrt{x-4}+3=\sqrt{x+11}$

Squaring both sides

$x-4+9+6\sqrt{x+4}=x+116\sqrt{x-4}=6\sqrt{x-4}=1$

Again squaring both sides

$x-4=1x=5$

So the values of $x$ are $\frac{1}{2}$ and $5$