Question

Solve the following equation:
$\sqrt{2x-3}+\sqrt{4x+1}=4$

Answer 1

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

$\Rightarrow 2x-3+4x+1+2\sqrt{2x+3}\sqrt{4x+1}=16$

${(6x-18)}^2=4(2x-3)(4x+1)$

$36(x^2+9-6x)=4[8x^2-12x+2x-3]$

$9x^2+81-54x=8x^2-10x-3$

$x^2+84-44x=0$

$\Rightarrow x^2-44x+84=0$

$\Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{44\pm \sqrt{{44}^2-4\times 1\times 84}}{2}$

$x=\frac{44\pm \sqrt{1600}}{2}$ or $x=\frac{44\pm 40}{2}$

$x=\frac{44+40}{2}=\frac{88}{2}=44$ or $\frac{44-40}{2}=\frac{4}{2}=2$

$\Rightarrow x=44$ or $2$

Hence, the answer is $44,2.$