If α and β(α<β) are the roots of the equation (-2+ 3 )(| x -3|)+(x-6 x )+(9-2 3 )=0, x ≥ 0 , then (β)/(α) + α β is equal to :

Correct answer: D. 10 Let t = x ≥ 0 . The equation becomes (-2+ 3 )|t-3| + (t^2 - 6t) + (9 - 2 3 ) = 0 . Since t^2 - 6t + 9 = (t-3)^2 , we get (t-3)^2 + ( 3 -2)|t-3| - 2 3 = 0 . Let u = |t-3| ≥ 0 : u^2 + ( 3 -2)u - 2 3 = 0 . Discriminant = ( 3 -2)^2 + 8 3 = 7 + 4 3 = (2+ 3 )^2 . u = (2- 3 ) ± (2+ 3 ) 2 , giving u = 2 or u = - 3 (rejected). |t-3| = 2 t = 5 or t = 1 , so x = 25 or x = 1 . α = 1, β = 25 . β/α + αβ = 25 + 25 = 5 + 5 = 10 .

JEE Main 202623 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 23 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Quadratic Equation

If α and β(α<β) are the roots of the equation (-2+ 3 )(| x...

\alpha

and

\beta(\alpha<\beta)

are the roots of the equation

(-2+\sqrt{3})(|\sqrt{x}-3|)+(x-6 \sqrt{x})+(9-2 \sqrt{3})=0, x \geqslant 0

, then

\sqrt{\frac{\beta}{\alpha}}+\sqrt{\alpha \beta}

is equal to :

Options

Correct answer

D. 10

Step-by-step solution

Let

t = \sqrt{x} \geq 0

. The equation becomes

(-2+\sqrt{3})|t-3| + (t^2 - 6t) + (9 - 2\sqrt{3}) = 0

.
Since

t^2 - 6t + 9 = (t-3)^2

, we get

(t-3)^2 + (\sqrt{3}-2)|t-3| - 2\sqrt{3} = 0

.
Let

u = |t-3| \geq 0

u^2 + (\sqrt{3}-2)u - 2\sqrt{3} = 0

.
Discriminant

= (\sqrt{3}-2)^2 + 8\sqrt{3} = 7 + 4\sqrt{3} = (2+\sqrt{3})^2

u = \frac{(2-\sqrt{3}) \pm (2+\sqrt{3})}{2}

, giving

u = 2

u = -\sqrt{3}

(rejected).

|t-3| = 2 \Rightarrow t = 5

t = 1

, so

x = 25

x = 1

\alpha = 1, \beta = 25

\sqrt{\beta/\alpha} + \sqrt{\alpha\beta} = \sqrt{25} + \sqrt{25} = 5 + 5 = 10

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