Question Id : 18078 |
Context : JEE Main 2026 (24 January Evening Shift)
The smallest positive integral value of $ a $, for which all the roots of $ x^4 - ax^2 + 9 = 0 $ are real and distinct, is equal to
🎥 Video solution / Text Solution of this question is given below:
$ x^4 - ax^2 + 9 = 0 \quad ...(1)$
let $ x^2 = t $
$ t^2 - at + 9 = 0 \quad ...(2)$
for roots of equation (1) to be real & distinct roots of equation (2) must be positive & distinct
(i) $ D > 0 \Rightarrow a^2 - 36 > 0 \Rightarrow a \in (-\infty, -6) \cup (6, \infty) $
(ii) $ \frac{-b}{2a} > 0 \Rightarrow \frac{a}{2} > 0 \Rightarrow a > 0 $
(iii) $ f(0) > 0 \Rightarrow 9 > 0 \Rightarrow a \in \mathbb{R} $
By (i) $\cap$ (ii) $\cap$ (iii)
$ \therefore a \in (6, \infty) $
$ \therefore $ least integral value of $ a $ is $ 7 $