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JEE Main 2007
Exam
JEE Main
Year
2007
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2007 · Mathematics · Quadratic Equation

If the difference between the roots of the equation x^2+a x+1=0 is...

If the difference between the roots of the equation x2+ax+1=0x^2+a x+1=0 is less than 5\sqrt{5}, then the set of possible values of aa is

Options

  1. A(−3,3)(-3,3)
  2. B(−3,∞)(-3, \infty)
  3. C(3,∞)(3, \infty)
  4. D(−∞,−3)(-\infty,-3)

Correct answer

A. (−3,3)(-3,3)

Step-by-step solution

x2+ax+1=0x^2+a x+1=0 α+β=−aαβ=1\alpha+\beta=-a \quad \alpha \beta=1 ∣α−β∣=(α+β)2−4αβ|\alpha-\beta|=\sqrt{(\alpha+\beta)^2-4 \alpha \beta} ∣α−β∣=a2−4|\alpha-\beta|=\sqrt{a^2-4} a2−4<5\sqrt{a^2-4} < \sqrt{5} a2−4<5a^2-4 < 5 a2−9<0a^2-9 < 0 a∈(−3,3)a \in(-3,3).
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