JEE Main 2026 Jan 23 Shift 1, Mathematics Q8: Some Other Forms of Equations

JEE Main2026Jan 23, Shift 1Mathematics

If $α$ and $β$ ( $α < β$ ) are the roots of the equation $(- 2 + 3) (∣ x - 3∣) + (x - 6 x) + (9 - 23) = 0$ , $x \geq 0$ , then $\frac{β}{α} + α β$ is equal to :

A
$8$
B
$9$
C
$10$
D
$11$

Solution

Rewrite using $∣ x - 3∣$ . Note $x - 6 x + 9 = (x - 3)^{2} = ∣ x - 3 ∣^{2}$ . Substituting:

$∣ x - 3 ∣^{2} - (2 - 3) ∣ x - 3∣ - 23 = 0$

Let $u = ∣ x - 3∣$ . Then $u^{2} - (2 - 3) u - 23 = 0$ , which factors as $(u - 2) (u + 3) = 0$ .

Since $u \geq 0$ , $u = 2$ , i.e. $∣ x - 3∣ = 2$ , giving $x = 1$ or $x = 5$ , so $x = 1$ or $x = 25$ .

$∴ α = 1$ , $β = 25$ . $β / α + α β = 5 + 5 = 10$ .

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JEE Main 2026 Jan 23 Shift 1, Mathematics Q8: Some Other Forms of Equations

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