Question

${\displaystyle \frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}}$ 

Answer 1

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

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

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

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

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

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

⇒${\displaystyle -9x^2+34x-8=0}$ 

⇒${\displaystyle 9x^2-34x-8=0}$ 

⇒${\displaystyle 9x(x-4)+2(x-4)=0}$ 

⇒${\displaystyle (x-4)(9x+2)=0}$ 

⇒${\displaystyle x-4=0 \text{or} 9x+2=0}$ 

⇒${\displaystyle x=4 \text{or} x=\frac{-2}{9}}$  

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