JEE Main202624 January 2026Evening ShiftMathematicsQuadratic EquationActual
The smallest positive integral value of a , for which all the roots of x⁴-a x²+9=0 are real and distinct, is equal to
Options
- A4
- B9
- C3
- D7
Correct answer
D. 7
Step-by-step solution
Let t = x^2 , so the equation becomes t^2 - at + 9 = 0 . For all four roots of the original equation to be real and distinct, both roots of this quadratic must be positive and distinct. Discriminant > 0 : a^2 - 36 > 0 |a| > 6 . Since a > 0 , we need a > 6 . Product of roots = 9 > 0 and sum of roots = a > 0 , so both roots are positive. The smallest positive integer satisfying a > 6 is a = 7 .