JEE Main 2025 · MathsMultiple choiceSingle correctEasyCalculation

JEE Main 7 April 2025, Shift 1, Maths Q3

Question 3 of 75 in this shift, Maths question 3 of 25, Section A.

Let the set of all values of

p \in \mathbb{R}

, for which both the roots of the equation

x^2 - (p+2)x + (2p+9) = 0

are negative real numbers, be the interval

(\alpha, \beta]

. Then

\beta - 2\alpha

is equal to

(1) $20$
(2) $9$
(3) $5$ Official answer
(4) $0$

Official answer

Option 3

NTA final key.

Chapter: Complex Numbers and Quadratic Equations
Topic: Location of Roots
Idea tested: Both Roots Negative

Same topic in other shifts

All Location of Roots questions

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6 Apr 2026, Shift 2 · Q16Let $\lim_{x\to2}\frac{(\tan(x-2))\left(rx^2+(p-2)x-2p\right)}{(x-2)^2}=5$ for some $r, p\in\mathbf{R}$ . If the set of all possible values…HardSingle correct
2 Apr 2025, Shift 2 · Q21If the set of all $a \in \mathbf{R} - \{1\}$ , for which the roots of the equation $(1-a)x^2 + 2(a-3)x + 9 = 0$ are positive is…MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.