Question

Solve the following quadratic equation by factorization, the roots are $3,1$
$\frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1},x\ne -1,\frac{1}{3}$

• A. True
• B. False

Answer 1

The correct option is A True

$\frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1}$

$\Rightarrow$ $\frac{6-x-1}{2x+2}=\frac{2}{3x-1}$

$\Rightarrow$ $\frac{5-x}{2x+2}=\frac{2}{3x-1}$

$\Rightarrow$ $(5-x)(3x-1)=2(2x+2)$

$\Rightarrow$ $15x-5-3x^2+x=4x+4$

$\Rightarrow$ $3x^2-12x+9=0$

$\Rightarrow$ $3(x^2-4x+3)=0$

$\Rightarrow$ $x^2-4x+3=0$

$\Rightarrow$ $x^2-3x-x+3=0$

$\Rightarrow$ $x(x-3)-1(x-3)=0$

$\Rightarrow$ $(x-3)(x-1)=0$

$\Rightarrow$ $x-3=0$ and $x-1=0$

$\therefore$ $x=3$ and $x=1$

$\Rightarrow$ Hence, we can see that roots given in question are correct.