Question

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

Answer 1

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

squaring both sides

$\Rightarrow {\left(\sqrt{2x+1}\right)}^2+{\left(\sqrt{x-3}\right)}^2+2\sqrt{2x+1}\sqrt{x-3}={\left(2\sqrt{x}\right)}^2$

$\Rightarrow 2x+1+x-3+2\sqrt{2x+1}\sqrt{x-3}=4x$

$\Rightarrow 2\sqrt{2x+1}\sqrt{x-3}=4x-3x+2=x+2$

squaring both sides

$\Rightarrow 4(2x+1)(x-3)={(x+2)}^2$

$4(2x^2+x-6x-3)=x^2+4+4x$

$8x^2+4x-24x-12=x^2+4+4x$

$7x^2-24x-16=0$

$\Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{24\pm \sqrt{{24}^2-4\times 7\times -16}}{2\times 7}$

$=\frac{24\pm \sqrt{1024}}{14}$

$\Rightarrow x=\frac{24\pm 4ac}{14}$

$\Rightarrow x=\frac{24+32}{14}=\frac{56}{14}=4$ or $x=\frac{24-32}{14}\frac{-8}{14}=\frac{-4}{7}$

Hence, solved.