# Solve each of the following quadratic equations:

Question:

Solve each of the following quadratic equations:

$\frac{x+3}{x-2}-\frac{1-x}{x}=4 \frac{1}{4}, x \neq 2,0$

Solution:

Given :

$\frac{x+3}{x-2}-\frac{1-x}{x}=4 \frac{1}{4}$

$\Rightarrow \frac{(x+3)}{(x-2)}-\frac{(1-x)}{x}=\frac{17}{4}$

$\Rightarrow \frac{x(x+3)-(1-x)(x-2)}{(x-2) x}=\frac{17}{4}$

$\Rightarrow \frac{x^{2}+3 x-\left(x-2-x^{2}+2 x\right)}{x^{2}-2 x}=\frac{17}{4}$

$\Rightarrow \frac{x^{2}+3 x+x^{2}-3 x+2}{x^{2}-2 x}=\frac{17}{4}$

$\Rightarrow \frac{2 x^{2}+2}{x^{2}-2 x}=\frac{17}{4}$

$\Rightarrow 8 x^{2}+8=17 x^{2}-34 x \quad[$ On cross multiplying $]$

$\Rightarrow-9 x^{2}+34 x+8=0$

$\Rightarrow 9 x^{2}-34 x-8=0$

$\Rightarrow 9 x^{2}-36 x+2 x-8=0$

$\Rightarrow 9 x(x-4)+2(x-4)=0$

$\Rightarrow(x-4)(9 x+2)=0$

$\Rightarrow x-4=0$ or $9 x+2=0$

$\Rightarrow x=4$ or $x=\frac{-2}{9}$

Hence, the roots of the equation are 4 and $\frac{-2}{9}$.