Question

Let $S$ be the set of all non-zero real numbers $\alpha$ such that the quadratic equation $\alpha x^2-x+a=0$ has two distinct real roots $x_1$ and $x_2$ satisfying the iequality $|x_1-x_2|<1$. 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. $(0,1)$
• D. $(\frac{1}{\sqrt{5}},\frac{1}{2})$

Answer 1

Answer: (C), (D)

$(\frac{1}{\sqrt{5}},\frac{1}{2})$
$(0,1)$

Given that

$\alpha x^2-x+a=0$ has distinct real roots.

Hence the discriminant of this quadratic equation should be greater than zero

If $a{x}^2+bx+c=0$ then $b^2-4ac>0$

Here, $a=\alpha ,b=-1,c=a$

$\Rightarrow (-1{)}^2-4\alpha a>0$

$\Rightarrow 1-4\alpha a>0$ $\left(1\right)$

Given that if $x_1,x_2$ are the roots, $|x_1-x_2|<1$

$\Rightarrow (x_1-x_2{)}^2<1$

We know that

$\Rightarrow (x_1-x_2{)}^2=(x_1+x_2{)}^2-4x_1{x}_2$

$\Rightarrow (x_1-x_2{)}^2=(\frac{1}{\alpha }{)}^2-\frac{4a}{\alpha }$

$\Rightarrow (x_1-x_2{)}^2=\frac{1-4a\alpha }{{\alpha }^2}$

Therefore, $\frac{1-4a\alpha }{{\alpha }^2}<1$

But Numerator is greater than Zero and denominator is always positive.

Hence, $0<\frac{1-4a\alpha }{{\alpha }^2}<1$

Therefore, $S=(0,1)$

Therefore, Option $C,D$ are a subset of $S$