Quadratic Equations – Location of Roots – JEE Main 24 Jan 2026 Shift 2

Solution:

Let $x^2 = t$. The equation becomes $t^2 – at + 9 = 0$.

For the original equation in $x$ to have 4 real and distinct roots, the quadratic in $t$ must have 2 distinct positive real roots ($t_1 > 0, t_2 > 0, t_1 \neq t_2$).

Conditions:
1. Discriminant $D > 0$:
$$a^2 – 4(1)(9) > 0 \Rightarrow a^2 – 36 > 0$$
$$(a-6)(a+6) > 0 \Rightarrow a \in (-\infty, -6) \cup (6, \infty)$$

2. Sum of roots $> 0$:
$$t_1 + t_2 = \frac{-(-a)}{1} = a > 0$$

3. Product of roots $> 0$:
$$t_1 t_2 = 9 > 0 \quad (\text{Always true})$$

Combining conditions 1 and 2:
$$a \in (6, \infty)$$

The smallest positive integer $a$ satisfying this is 7.

Ans. (4)