Question

Solve the following quadratic equations by factorization method :
$\frac{4}{x}-3=\frac{5}{2x+3}$

Answer 1

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

$⟹\frac{4-3x}{x}=\frac{5}{2x+3}$

$⟹(4-3x)(2x+3)=5x$

$⟹8x+12-6x^2-9x=5x$

$⟹-6x^2-6x-12=0$

$⟹x^2-x-2=0$

$a=1,b=-1,c=-2$

$D=b^2-4ac$

$⟹D=(-1{)}^2-4(1\left)\right(-2)$

$⟹D=1+8$

$⟹D=9$

$\therefore$ Roots are real and distinct.