Q78 / 120 · JEE Main 2007 · Mathematics

If the difference between the roots of the equation $x^2+a x+1=0$ is less than $\sqrt{5}$, then the set of possible values of $a$ is

  • Correct answer
    A
    $(-3,3)$
  • B
    $(-3, \infty)$
  • C
    $(3, \infty)$
  • D
    $(-\infty,-3)$

Solution

$x^2+a x+1=0$ $\alpha+\beta=-a \quad \alpha \beta=1$ $|\alpha-\beta|=\sqrt{(\alpha+\beta)^2-4 \alpha \beta}$ $|\alpha-\beta|=\sqrt{a^2-4}$ $\sqrt{a^2-4} < \sqrt{5}$ $a^2-4 < 5$ $a^2-9 < 0$ $a \in(-3,3)$.

Hence, the correct answer is option: 'A'.

Relevant Questions