Mathematics · Quadratic Equations

JEE Main 2024 — 4 April, Shift 1 — Question 14

If 2 and 6 are the roots of the equation $a x^{2}+b x+1=0$ , then the quadratic equation, whose roots are $\frac{1}{2 a+b}$ and $\frac{1}{6 a+b}$ , is :

Option A:
$2 x^{2}+11 x+12=0$
Option B:
$4 x^{2}+14 x+12=0$
Option C:
$x^{2}+10 x+16=0$
Option D:
$x^{2}+8 x+12=0$
Correct

Answer: D

Step-by-step solution

Sum $=8=-\frac{\mathrm{b}}{\mathrm{a}}$

$\text { Product }=12=\frac{1}{\mathrm{a}}$

$\Rightarrow \mathrm{a} =\frac{1}{12}$ , $\mathrm{~b} =-\frac{2}{3}$

$2 \mathrm{a}+\mathrm{b}=\frac{2}{12}-\frac{2}{3}=-\frac{1}{2}$

$6 a+b=\frac{6}{12}-\frac{2}{3}=-\frac{1}{6}$

sum $=-8$

$\mathrm{P}=12$

$\mathrm{x}^{2}+8 \mathrm{x}+12=0$

Answer key and solution verified before publishing.

Practise Quadratic Equations

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Mathematics
Chapter: Quadratic Equations
Topic: Theory of Quadratic Equations