Question

After solving this equation $\frac{x^2+8x-3}{8x-3}=\frac{x^2+4x+4}{4x+4}$, $x$ is either $\frac{7}{4}$ or $m$. Then, the value of $m$ is:

Answer 1

$\frac{x^2+8x-3}{8x-3}=\frac{x^2+4x+4}{4x+4}$

$\Rightarrow$ $(x^2+8x3)(4x+4)=\left(8x3\right)(x^2+4x+4)$

$\Rightarrow$$4x^3+36x^2+20x-12=8x^3+29x^2+20x-12$

$\Rightarrow$$x^2(4x+36)=x^2(8x+29)$

$\Rightarrow$$x=0,4x+36=8x+29$

$\Rightarrow$$4x=7$

$\Rightarrow$$x=0,\frac{7}{4}$

Hence, $m=0$.