Question

Can someone help me solve this problem asap

Answer 1

View attachment :) hope i helped

Answer 2

Hello,

We suppose x≠-2.

$\dfrac{x^2+5x-2}{x+2}=3\\\\ ==\textgreater \ \dfrac{x^2+5x-2-3(x+2)}{x+2}=0\\\\ ==\textgreater \ x^2+4x-2x-8=0\\ ==\textgreater \ x^2+4x-2x-8=0\\ ==\textgreater \ x(x+4)-2(x+4)=0\\ ==\textgreater \ (x+4)(x-2)=0\\ ==\textgreater \ x=-4 \ or \ x=2\\\\ Proof:\\ if\ x=-4\ then \ \ \dfrac{(-4)^2+5*(-4)-2}{-4+2} =\frac{16-20-2}{-2}=\frac{-6}{-2}=3\\\\ if\ x=2\ then \ \ \dfrac{(2)^2+5*(2)-2 }{2+2} =\frac{4+10-2}{4}=\frac{12}{4}=3$