JEE Main 2026 — Mathematics Algebra

The given quadratic equation is $x^2 - ax + 2 = 0$x2−ax+2=0 with roots $e_1$e1​ and $e_2$e2​.

The sum of the roots is $e_1 + e_2 = a$e1​+e2​=a and the product is $e_1 e_2 = 2$e1​e2​=2.

For the roots to be real and distinct, the discriminant must be positive:

$$\Delta = a^2 - 8 > 0 \Rightarrow a \in (-\infty, -2\sqrt{2}) \cup (2\sqrt{2}, \infty)$$Δ=a2−8>0⇒a∈(−∞,−22​)∪(22​,∞)

Since $e_1$e1​ and $e_2$e2​ represent eccentricities, they must be positive. Given $e_1 e_2 = 2 > 0$e1​e2​=2>0, both roots have the same sign. For them to be positive, their sum must be positive, so $a > 0$a>0. Thus, $a > 2\sqrt{2}$a>22​.

For $e_1$e1​ and $e_2$e2​ to be the eccentricities of hyperbolas, both must be strictly greater than $1$1. This requires $1$1 to lie outside the roots and the vertex of the parabola to be to the right of $1$1:

$$(e_1 - 1)(e_2 - 1) > 0 \Rightarrow e_1 e_2 - (e_1 + e_2) + 1 > 0$$(e1​−1)(e2​−1)>0⇒e1​e2​−(e1​+e2​)+1>0

$$2 - a + 1 > 0 \Rightarrow a < 3$$2−a+1>0⇒a<3

Combining this with the discriminant condition, we get $a \in (2\sqrt{2}, 3)$a∈(22​,3).

Thus, $\alpha = 2\sqrt{2}$α=22​ and $\beta = 3$β=3.

For $e_1$e1​ and $e_2$e2​ to be the eccentricities of an ellipse and a hyperbola, one root must be in $(0, 1)$(0,1) and the other in $(1, \infty)$(1,∞). This requires $1$1 to lie between the roots:

$$(e_1 - 1)(e_2 - 1) < 0 \Rightarrow 2 - a + 1 < 0 \Rightarrow a > 3$$(e1​−1)(e2​−1)<0⇒2−a+1<0⇒a>3

For $a > 3$a>3, the roots are real, distinct, and positive. Thus, the set of such $a$a is $(3, \infty)$(3,∞).

This gives $\gamma = 3$γ=3.

Finally, calculating the required sum:

$$\alpha^2 + \beta^2 + \gamma^2 = (2\sqrt{2})^2 + 3^2 + 3^2 = 8 + 9 + 9 = 26$$α2+β2+γ2=(22​)2+32+32=8+9+9=26

Answer: $26$26