Question

Let $S$ be the set of all non-zero real numbers $\alpha$ such that the quadratic equation $\alpha x^2-x+\alpha =0$ has two distinct real roots $x_1$ and $x_2$ satisfying the inequality $|x_1-x_2|<1.$ Which of the following intervals is(are) subset(s) of S?

• A. $(\frac{-1}{2},\frac{-1}{\sqrt{5}})$
• B. $(\frac{-1}{\sqrt{5}},0)$
• C. $(0,\frac{1}{\sqrt{5}})$
• D. $(\frac{1}{\sqrt{5}},\frac{1}{2})$

Answer 1

Answer: (A), (D)

The correct options are
A $(\frac{-1}{2},\frac{-1}{\sqrt{5}})$
D $(\frac{1}{\sqrt{5}},\frac{1}{2})$

Given, quadratic equation $\alpha x^2-x+\alpha =0$ has two distinct real roots $x_1$ and $x_2$ satisfying the inequality $|x_1-x_2|<1$.

$x_1+x_2=\frac{1}{\alpha }$ and $x_1{x}_2=1$

$|x_1-x_2|=|\sqrt{(x_1+x_2{)}^2-4x_1{x}_2}|=|\sqrt{\frac{1}{{\alpha }^2}-4}|<1$

$\Rightarrow 0<\frac{1}{{\alpha }^2}-4<1$

$\Rightarrow 4<\frac{1}{{\alpha }^2}<5$

$\Rightarrow {\alpha }^2>\frac{1}{5}$ and ${\alpha }^2<\frac{1}{4}$

$\therefore \alpha \in \left(\frac{-1}{2},\frac{-1}{\sqrt{5}}\right)\cup \left(\frac{1}{\sqrt{5}},\frac{1}{2}\right)$

Hence, options A and D.