Question

The number of distinct real roots of the equation $\bigg(\frac{x+1}{x}\bigg)^2-3\bigg(\frac{x+1}{x}\bigg)+2=0$ equals

Answer 1

Let $\frac{x+1}{x} = a$

The given equation becomes, $a^2-3a+2 = 0$ $a= 2$ or $1$ i.e. $\frac{x+1}{x}$ = $2$ or $\frac{x+1}{x} = 1$

since $x$ is real, $x+\frac{1}{x} ≠1$;

∴ $\frac{x+1}{x} = 2$

$∴$ The number of solutions = $1$