JEE Main 2026 Jan 24 Shift 2, Mathematics Q18: Quadratic Equations

JEE Main2026Jan 24, Shift 2Mathematics

The smallest positive integral value of $a$ , for which all the roots of $x^{4} - a x^{2} + 9 = 0$ are real and distinct, is equal to

Solution

Let $t = x^{2}$ . Then $t^{2} - a t + 9 = 0$ . For the original quartic to have 4 real distinct roots, this quadratic in $t$ must have two distinct positive real roots.

(i) Discriminant $> 0$ : $a^{2} - 36 > 0 \Rightarrow a > 6$ or $a < - 6$ .

(ii) Sum of roots $> 0$ : $a > 0$ .

(iii) Product of roots $> 0$ : $9 > 0$ . ✓

Combining: $a > 6$ . Smallest positive integer is $a = 7$ .

Concept behind this question

Quadratic EquationsNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2026 Jan 24 Shift 2, Mathematics Q18: Quadratic Equations

Concept behind this question

More previous year questions on Quadratic Equations

Practice more Mathematics