Question

Solve the following equations:
$\sqrt{2x^2+5x-7}+\sqrt{3(x^2-7x+6)}-\sqrt{7x^2-6x-1}=0$.

Answer 1

$\sqrt{2x^2+5x-7}+\sqrt{{3(x}^2-7x+6)}-\sqrt{7x^2-6x-1}=0\Rightarrow \sqrt{2x^2-2x+7x-7}+\sqrt{{3(x}^2-x-6x+6)}-\sqrt{7x^2-7x+x-1}=0\Rightarrow \sqrt{2x(x-1)+7(x-1)}+\sqrt{3\left\{x\right(x-1)-6(x-1\left)\right\}}-\sqrt{7x(x-1)+1(x-1)}=0\Rightarrow \sqrt{(2x+7)(x-1)}+\sqrt{3(x-6)(x-1)}-\sqrt{(7x+1)(x-1)}=0\Rightarrow \sqrt{x-1}(\sqrt{2x+7}+\sqrt{3x-18}-\sqrt{7x+1})=0$

$\therefore \sqrt{x-1}=0$ or, $\sqrt{2x+7}+\sqrt{3x-18}-\sqrt{7x+1}=0$

$\Rightarrow x=1$

Also $\sqrt{2x+7}+\sqrt{3x-18}-\sqrt{7x+1}=0$

$\sqrt{2x+7}+\sqrt{3x-18}=\sqrt{7x+1}$

Squaring both sides

$\Rightarrow 2x+7+3x-18+2\sqrt{2x+7}\sqrt{3x-18}=7x+1\Rightarrow 2\sqrt{2x+7}\sqrt{3x-18}=2x+12\Rightarrow \sqrt{2x+7}\sqrt{3x-18}=x+6$

Again squaring both sides

$\Rightarrow \sqrt{2x+7}\sqrt{3x-18}=x+6\Rightarrow (2x+7)(3x-18)={(x+6)}^2\Rightarrow {6x}^2-15x-126=x^2+36+12x\Rightarrow {5x}^2-27x-162=0\Rightarrow {5x}^2-45x+18x-162=0\Rightarrow 5x(x-9)+18(x-9)=0\Rightarrow (5x+18)(x-9)=0\Rightarrow x=-\frac{18}{5},9$

So the values of $x$ are $-\frac{18}{5},9$ and $1$