Question

Solve the following equation.
$\frac{x^2+12x+20}{3x-5}=\frac{x^2+8x+12}{2x+3}$.

Answer 1

Consider the given equation.

$\frac{x^2+12x+20}{3x-5}=\frac{x^2+8x+12}{2x+3}$

$\frac{x^2+10x+2x+20}{3x-5}=\frac{x^2+6x+2x+12}{2x+3}$

$\frac{x(x+10)+2(x+10)}{3x-5}=\frac{x(x+6)+2(x+6)}{2x+3}$

$\frac{(x+10)(x+2)}{(3x-5)}=\frac{(x+6)(x+2)}{(2x+3)}$

$\frac{(x+10)}{(3x-5)}=\frac{(x+6)}{(2x+3)}$

$(x+10)(2x+3)=(x+6)(3x-5)$

$2x^2+3x+20x+30=3x^2-5x+18x-30$

$2x^2+23x+30=3x^2+13x-30$

$x^2-10x-60=0$

$x=\frac{10\pm \sqrt{100-4\times 1\times -60}}{2}$

$x=\frac{10\pm \sqrt{100+240}}{2}$

$x=\frac{10\pm \sqrt{340}}{2}$

$x=\frac{10\pm 2\sqrt{85}}{2}$

$x=5\pm \sqrt{85}$

Hence, the value of $x$ is $5\pm \sqrt{85}$.