Question

Solve the following inequality:
$\frac{x^2-36}{x^2-9x+18}<0$

Answer 1

Given equation is $\frac{x^2-36}{x^2-9x+18}$<0

Now solve for equation for equal to zero. now we have,

$x^2-36=0$ but $x^2-9x+18=0$

now $x^2-36=0$

$x^2=36$

$-6<x<6$

Now, $x^2-9x+18=0$

$x^2-6x-3x+18=0$

$x(x-6)-3(x-6)=0$

$(x-6)(x-3)=0$

$x\in (6,3)$

so the final value for x for the equation $\frac{x^2-36}{x^2-9x+18}<0$ is

$x\in (-6,3)$