Question

Solve the following quadratic equation by factorization, the roots are $0,-7$
$3\left(\frac{3x-1}{2x+3}\right)-2\left(\frac{2x+3}{3x-1}\right)=5;x\ne \frac{1}{3},-\frac{3}{2}$

• A. True
• B. False

Answer 1

The correct option is A True

$3\left(\frac{3x-1}{2x+3}\right)-2\left(\frac{2x+3}{3x-1}\right)=5$

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

$\Rightarrow$ $3p-\frac{2}{p}=5$

$\Rightarrow$ $3p^2-2=5p$

$\Rightarrow$ $3p^2-5p-2=0$

$\Rightarrow$ $3p^2-6p+p-2=0$

$\Rightarrow$ $3p(p-2)+1(p-2)=0$

$\Rightarrow$ $(p-2)(3p+1)=0$

$\Rightarrow$ $p-2=0$ and $3p+1=0$

$\therefore$ $p=2$ and $p=-\frac{1}{3}$

Re-substituting value of $p$,

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

$\Rightarrow$ $3x-1=4x+6$ and $9x-3=-2x-3$

$\therefore$ $x=-7$ and $x=0$

$\therefore$ We can see roots given in question are correct.