Mathematics · Quadratic Equations

JEE Main 2025 — 23 January, Morning Shift — Question 24

If the set of all values of $a$ , for which the equation $5 x^{3}-15 x-a=0$ has three distinct real roots, is the interval $(\alpha, \beta)$ , then $\beta-2 \alpha$ is equal to $\_\_\_\_\_$

Answer: 30

Numerical answer — enter this value.

Step-by-step solution

$5 \mathrm{x}^{3}-15 \mathrm{x}-\mathrm{a}=0$

\begin{aligned} & f(x)=5 x^{3}-15 x \\& f^{\prime}(x)=15 x^{2}-15=15(x-1)(x+1) \end{aligned}

$\mathrm{a} \in(-10,10)$

$\alpha=-10, \beta=10$

$\beta-2 \alpha=10+20=30$

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Exam: JEE Main 2025
Paper: 23 January, Morning Shift
Subject: Mathematics
Chapter: Quadratic Equations
Topic: Location of roots of a Quadratic Equation

If the equation a(b-c) x^2+b(c-a) x+c(a-b)=0 has equal roots, where a+ c=15 and b=36/5 , then a^2+ c^2 is equal to \ \ \ \ \

Answer: 30

Step-by-step solution

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