Question

Solve each of the following quadratic equations:

$\frac{16}{x}-1=\frac{15}{x+1},x\ne 0,-1$ [CBSE 2014]

Answer 1

$$\begin{gathered} \frac{16}{x}-1=\frac{15}{x+1},x\ne 0,-1 \\ \Rightarrow \frac{16}{x}-\frac{15}{x+1}=1 \\ \Rightarrow \frac{16x+16-15x}{x\left(x+1\right)}=1 \\ \Rightarrow \frac{x+16}{x^2+x}=1 \end{gathered}$$
$$\begin{gathered} \Rightarrow x^2+x=x+16\left(\mathrm{Cross}\mathrm{multiplication}\right) \\ \Rightarrow x^2-16=0 \\ \Rightarrow \left(x+4\right)\left(x-4\right)=0 \\ \Rightarrow x+4=0\mathrm{or}x-4=0 \end{gathered}$$
$\Rightarrow x=-4\mathrm{or}x=4$
Hence, −4 and 4 are the roots of the given equation.