Aspire Faculty ID #18078 · Topic: JEE Main 2026 (24 January Evening Shift) · Just now
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

Solution

$ 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 $

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