Question

Solve the following equation:
$\sqrt{3-x}+\frac{6}{\sqrt{3-x}}=\sqrt{9-5x}$

Answer 1

$\sqrt{3-x}+\frac{6}{\sqrt{3-x}}=\sqrt{9-5x}$

squaring both sides, we get

$\Rightarrow 3-x+\frac{36}{3-x}+2\times 6\times \sqrt{3-x}\times \frac{1}{\sqrt{3-x}}=9-5x$

$\Rightarrow 3-x+\frac{36}{3-x}+12=9-5x$

$\Rightarrow {(3-x)}^2+36+12(3-x)=(9-5x)(3-x)$

$\Rightarrow x^2+9-6x+36+36-12x=27-15x-9x+5x^2$

$\Rightarrow x^2-18x+81=-24x+5x^2=27$

$\Rightarrow 5x^2-x^2-24x+18x+27-81=0$

$\Rightarrow 4x^2-6x-54=0$

$\Rightarrow 2x^2-3x-27=0$

$\Rightarrow x=\frac{3\pm \sqrt{9-4\times 2\times -27}}{2\times 2}$

$x=\frac{3\pm \sqrt{225}}{4}$ $\frac{3\pm \sqrt{225}}{4}$

$x=\frac{3+15}{4}$ $x=\frac{3-15}{2}$

$x=\frac{18}{4}=\frac{9}{2}$ $x=\frac{-12}{2}=-6$

$\therefore x=-6,\frac{9}{2}$

Hence, solved.