If the difference between the roots of x²+ax+1=0 is less than 5, then the set of possible values of a is (a) (3,∞) (b) (-∞,-3) (c) (-3,3) (d) (-3,∞)

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[JEE Main 2007] if the difference between the roots of x 2 ax 1 0 is less

VAVidaara Admin Asked 2d ago 0 views 1 answer

If the difference between the roots of $x^2+ax+1=0$ is less than $\sqrt5$, then the set of possible values of $a$ is

(a) $(3,\infty)$
(b) $(-\infty,-3)$
(c) $(-3,3)$
(d) $(-3,\infty)$

VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 2d ago ▲ 0

Correct answer: (c) $(-3,3)$

$(\text{diff})^2=a^2-4<5\Rightarrow a^2<9\Rightarrow-3<a<3$.

JEE Main 2007 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.

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