JEE Mathematics 2026 Question with Solution

Given: $x^4-ax^2+9=0$

Find: The smallest positive integral value of $a$ such that all roots are real and distinct.

Substitute

$$t=x^2$$

Then the equation becomes

$$t^2-at+9=0$$

For the roots of this quadratic in $t$ to be real and distinct, its discriminant must be positive:

$$a^2-36>0$$

So,

$$a>6$$

since $a$ is positive.

Also, the roots of $t$ must be positive so that $x^2=t$ gives real values of $x$. Here,

$$\text{sum of roots}=a>0, \qquad \text{product of roots}=9>0$$

Therefore both roots of $t$ are positive.

Since the two positive roots of $t$ are distinct, each gives two distinct real roots of $x$, so all four roots are real and distinct.

Hence the smallest positive integer value is $7$.

Therefore, the correct option is C.