Question

After solving $\frac{x^2-5x+12}{5x-12}=\frac{x^2-x-4}{x+4}$ for $(x\ge 0)$, $x$ is $\{0,m\}$. Then, the value of $m$ is :

Answer 1

$\frac{x^2-5x+12}{5x-12}=\frac{x^2-x-4}{x+4}$

$\Rightarrow$ $(x^2-5x+12)(x+4)=(5x-12)(x^2-x-4)$

$\Rightarrow$ $x^3-x^2-8x+48=5x^3-17x^2-8x+48$

$\Rightarrow$ $x^2(x-1)=x^2(5x-17)$

$\Rightarrow$ $x=0,x-1=5x-17$

$\Rightarrow$ $x=0,4x=16$

$\Rightarrow$ $x=0,4$

According to the question, $x$ is $0,m$

Hence, $m=4$