Let α, β be the roots of the equation x 2 +2 √(2) x-1=0 . The quadratic equation, whose roots are α 4 +β 4 and 1/10 (α 6 +β 6 ) , is :

Mathematics · Quadratic Equations

JEE Main 2024 — 9 April, Shift 1 — Question 19

Let $\alpha, \beta$ be the roots of the equation $x^{2}+2 \sqrt{2} x-1=0$ . The quadratic equation, whose

roots are $\alpha^{4}+\beta^{4}$ and $\frac{1}{10}\left(\alpha^{6}+\beta^{6}\right)$ , is :

Option A:
$x^{2}-190 x+9466=0$
Option B:
$x^{2}-195 x+9466=0$
Option C:
$x^{2}-195 x+9506=0$
Correct
Option D:
$x^{2}-180 x+9506=0$

Answer: C

Step-by-step solution

$x^{2}+2 \sqrt{2} x-1=0$

$\alpha+\beta=-2 \sqrt{2}$

$\alpha \beta=-1$

$\alpha^{4}+\beta^{4}=\left(\alpha^{2}+\beta^{2}\right)^{2}-2 \alpha^{2} \beta^{2}$

$=\left((\alpha+\beta)^{2}-2 \alpha \beta\right)^{2}-2(\alpha \beta)^{2}$

$=(8+2)^{2}-2(-1)^{2}$ $=100-2=98$

$\alpha^{6}+\beta^{6}=\left(\alpha^{3}+\beta^{3}\right)^{2}-2 \alpha^{3} \beta^{3}$ $=\left((\alpha+\beta)\left((\alpha+\beta)^{2}-3 \alpha \beta\right)^{2}-2(\alpha \beta)^{3}\right.$

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

$=(8)(121)+2=970$

$\frac{1}{10}\left(\alpha^{6}+\beta^{6}\right)=97$

$x^{2}-(98+97) x+(98)(97)=0$

$\Rightarrow \mathrm{x}^{2}-195 \mathrm{x}+9506=0$

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: 9 April, Shift 1
Subject: Mathematics
Chapter: Quadratic Equations
Topic: Theory of Quadratic Equations