Question:
Solve each of the following quadratic equations:
$\frac{16}{x}-1=\frac{15}{x+1}, x \neq 0,-1$
Solution:
$\frac{16}{x}-1=\frac{15}{x+1}, x \neq 0,-1$
$\Rightarrow \frac{16}{x}-\frac{15}{x+1}=1$
$\Rightarrow \frac{16 x+16-15 x}{x(x+1)}=1$
$\Rightarrow \frac{x+16}{x^{2}+x}=1$
$\Rightarrow x^{2}+x=x+16 \quad$ (Cross multiplication)
$\Rightarrow x^{2}-16=0$
$\Rightarrow(x+4)(x-4)=0$
$\Rightarrow x+4=0$ or $x-4=0$
$\Rightarrow x=-4$ or $x=4$
Hence, −4 and 4 are the roots of the given equation.
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