Verified Solution • 2025–26 Session moderate

QJEEM07Ma2f74: JEE Main 2007 Mathematics - Class 11 JEEM07Ma2f74 Solution: If the difference between the roots of the equation is less…

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

Multiple Choice Options

  • A (3,∞)\left( {3,\infty } \right)
  • B (−∞,−3)\left( { - \infty , - 3} \right)
  • C (−3,3)\left( { - 3,3} \right)
  • D (−3,∞)\left( { - 3,\infty } \right)

Step-by-Step Verified Solution

Let α\alpha and β\beta are roots of the equation x2+ax+1=0{x^2} + ax + 1 = 0

So, α+β=−a\alpha + \beta = - a and αβ=1\alpha \beta = 1

given \left| {\alpha - \beta } \right| < \sqrt 5

\Rightarrow \sqrt {{{\left( {\alpha - \beta } \right)}^2} - 4\alpha \beta } < \sqrt 5

(as (α−β)2=(α+β)2−4αβ{\left( {\alpha - \beta } \right)^2} = {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta )

\Rightarrow \sqrt {{a^2} - 4} < \sqrt 5

\Rightarrow {a^2} - 4 < 5

\Rightarrow {a^2} - 9 < 0 \Rightarrow {a^2} < 9

\Rightarrow - 3 < a < 3

⇒a∈(−3,3)\Rightarrow a \in \left( { - 3,3} \right)

Related Learning Resources & Exam Preparation