Question

Solve the following quadratic equation by factorization , the roots are:$-2,1$
$\frac{4}{x}-3=\frac{5}{2x+3},x\ne 0,-\frac{3}{2}$

• A. True
• B. False

Answer 1

The correct option is B False

$\frac{4}{x}-3=\frac{5}{2x+3}$

$\Rightarrow$ $\frac{4-3x}{x}=\frac{5}{2x+3}$

$\Rightarrow$ $(4-3x)(2x+3)=5x$

$\Rightarrow$ $8x+12-6x^2-9=5x$

$\Rightarrow$ $5x-8x-12+6x^2+9=0$

$\Rightarrow$ $6x^2-3x-3=0$

$\Rightarrow$ $3(2x^2-x-1)=0$

$\Rightarrow$ $2x^2-x-1=0$

$\Rightarrow$ $2x^2-2x+x-1=0$

$\Rightarrow$ $2x(x-1)+1(x-1)=0$

$\Rightarrow$ $(x-1)(2x+1)=0$

$\Rightarrow$ $x-1=0$ and $2x+1=0$

$\therefore$ $x=1$ and $x=\frac{-1}{2}$

$\therefore$ We can see, roots given question are incorrect.