Question

Solve $\sqrt{24-10x}=3-4x$

Answer 1

Given $\sqrt{24-10x}=3-4x$
Squaring on both sides, we get
$(\sqrt{24-10x}{)}^2=(3-4x{)}^{2}24-10x=9-24x+16x^{2}16x^2-24x+10x+9-24=0$
$\therefore 16x^2-24x+10x-15=08x(2x-3)+5(2x-3)=0$
$\therefore (2x-3)(8x+5)=0$

$2x-3=0$ or $8x+5=0$
$\therefore 2x=3$ or $8x=-5,x=\frac{3}{2}$ or $x=\frac{-5}{8}$
$\therefore \frac{3}{2}$ and $\frac{-5}{8}$ are the roots of $\sqrt{24-10x}=3-4x$.