Mathematics · Quadratic Equations

JEE Main 2025 — 7 April, Morning Shift — Question 23

Let the set of all values $p \in \mathbb{R}$ , for which both the roots of the equation $x^{2}-(p+2) x+(2 p+9)=0$

are negative real numbers, be the interval $(\alpha, \beta]$ . Then $\beta-2 \alpha$ is equal to

Answer: B

$x^{2}-(p+2) x+(2 p+9)=0$

$D \geq 0$

$(p+2)^{2}-4(2 p+9) \geq 0$

$p^{2}+4 p+4-8 p-36 \geq 0$

$p^{2}-4 p-32 \geq 0$

$(p-8)(p+4) \geq 0$

$p \in(-\infty,-4] \cup[8, \infty) …(1)$

Sum: $p+2<0$

$p<-2 …(2)$

Product $>0$

$2 p+9>0$

$p>\frac{-9}{2} …(3)$

From (1), (2) and (3) $p \in\left(\frac{-9}{2},-4\right]$

$\beta-2 \alpha=-4-2\left(\frac{-9}{2}\right)$

$=5$

Answer key and solution verified before publishing.

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