Question

If $\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15},x\ne 0$ and $x\ne -1$, then value of $x$ is

• A. $3,4$
• B. $\frac{-5}{2},\frac{3}{2}$
• C. $\frac{5}{2},\frac{3}{2}$
• D. $0$

Answer 1

The correct option is B $\frac{-5}{2},\frac{3}{2}$

Given equation is: $\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15}$

$\Rightarrow \frac{x^2+{(x+1)}^2}{x(x+1)}=\frac{34}{15}$
$\Rightarrow \frac{x^2+x^2+2x+1}{x^2+x}=\frac{34}{15}$

$\Rightarrow \frac{{2x}^2+2x+1}{x^2+x}=\frac{34}{15}$

$\Rightarrow 15({2x}^2+2x+1)=34(x^2+x)$

$\Rightarrow 30x^2+30x+1=34x^2+34x$

$\Rightarrow 4x^2+10x-6x-15=0$
$\Rightarrow 2x(2x+5)-3(2x+5)=0$
$\Rightarrow (2x+5)(2x-3)=0$
$\Rightarrow 2x+5=0$ and $2x-3=0$

$\Rightarrow x=-\frac{5}{2}$ and $x=\frac{3}{2}$
Hence, $x=-\frac{5}{2},\frac{3}{2}$ are the required solutions.