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 :

Correct answer: C. x^2-195 x+9506=0 aligned & x^2+2 2 x-1=0 & α+β=-2 2 & α β=-1 & α^4+β^4=(α^2+β^2)^2-2 α^2 β^2 & =((α+β)^2-2 α β)^2-2(α β)^2 & =(8+2)^2-2(-1)^2 & =100-2=98 & α^6+β^6=(α^3+β^3)^2-2 α^3 β^3 & =((α+β)((α+β)^2-3 α β)^2-2(α β)^3. aligned aligned & =(-2 2 (8+3))^2+2 & =(8)(121)+2=970 & (1)/(10)(α^6+β^6)=97 & x^2-(98+97) x+(98)(97)=0 & x^2-195 x+9506=0 aligned

JEE Main 20249 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Quadratic Equation

Let α, β be the roots of the equation x^2+2 2 x-1=0 . The quadratic...

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 :

Options

A $x^2-190 x+9466=0$
B $x^2-180 x+9506=0$
C $x^2-195 x+9506=0$
D $x^2-195 x+9466=0$

Correct answer

C. $x^2-195 x+9506=0$

Step-by-step solution

\begin{aligned} & 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.\end{aligned}

\begin{aligned} & =(-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 x^2-195 x+9506=0\end{aligned}

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