Solve the following quadratic equation by factorization method:

Q) Solve the following quadratic equation by factorization method: $\frac{4}{x}-3=\frac{5}{2x+3x},~x\neq 0,\frac{-3}{2}$

Ans:

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

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

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

$12-x-6x^{2}=5x$

$6x^{2}+6x-12=0$

$x^{2}+x-2=0$

$x^{2}+2x-x-2=0$

$x(x+2)-(x+2)=0$

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

$⇒ x + 2 = 0 ~and ~x-1=0$

$x=-2~and~ x=1$

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