Question

Let $S$ be the set of all non-zero real number $\alpha$ 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 interval is(are) a subset 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)

$(\frac{1}{\sqrt{5}},\frac{1}{2})$

Given quadratic equation
$\alpha x^2-x+\alpha =0$ has two distinct real roots $x_1$ and $x_2$
$\therefore D>0\Rightarrow 1-4{\alpha }^2>0\Rightarrow -\frac{1}{2}<\alpha <\frac{1}{2}\cdots \left(1\right)$

Now,
$x_1+x_2=\frac{1}{\alpha }\text{and}{x}_1\cdot x_2=1$
Since
$|x_1-x_2|<1\Rightarrow (x_1-x_2{)}^2<1\Rightarrow (x_1+x_2{)}^2-4x_1{x}_2<1\Rightarrow \frac{1}{{\alpha }^2}-4<1\Rightarrow 5{\alpha }^2-1>0\Rightarrow (\sqrt{5}\alpha -1)(\sqrt{5}\alpha +1)>0$
$\alpha <-\frac{1}{\sqrt{5}}\text{and}\alpha >\frac{1}{\sqrt{5}}\cdots \left(2\right)$

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