Question

The number of integral values of $a$ such that the quadratic equation $4a{x}^2+5x+a=0$ has two distinct real roots $x_1$ and $x_2$, satisfying the inequality $|x_1-x_2|<1$, is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

$4a{x}^2+5x+a=0$
$|x_1-x_2|<1$
$\Rightarrow \left|\frac{\sqrt{25-16a^2}}{4a}\right|<1$
$\Rightarrow \sqrt{25-16a^2}<|4a|$
$\Rightarrow a^2>\frac{25}{32}$
$\Rightarrow a\in (-\infty ,-\frac{5}{4\sqrt{2}})\cup (\frac{5}{4\sqrt{2}},\infty )...\left(1\right)$

Also, the roots are real and distinct.
So, $25-16a^2>0$
$\Rightarrow a^2<\frac{25}{16}$
$\Rightarrow a\in (-\frac{5}{4},\frac{5}{4})...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$a\in (-\frac{5}{4},-\frac{5}{4\sqrt{2}})\cup (\frac{5}{4\sqrt{2}},\frac{5}{4})$
Possible integral values of $a$ are $-1,1$
Hence, the number of integral values of $a$ is $2$