Let α, β be the roots of the equation x^2-a x-b=0 with Im(α) Im(β) . Let P_n=α^n-β^n . If P_3=-5 7 i, P_4=-3 7 i, P_5=11 7 i and P_6=45 7 i , then |α^4+β^4| is equal to .

Correct answer: 0 aligned & α+β=a α β=-b & P_6=aP_5+bP_4 & 45 7 i=a × 11 7 i+b(-3 7 ) i & 45=11 a-3 ~b aligned and aligned & P_5=aP_4+b P_3 & 11 7 i=a(-3 7 i)+b(-5 7 i) & 11=-3 a-5 ~b & a=3, ~b=-4 & |α^4+β^4|= (α^4-β^4)^2+4 α^4 β^4 & = -63+4.4^4 & = -63+1024 = 961 =31 aligned

JEE Main 202523 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2025 · 23 January · Evening Shift · Mathematics · Complex Number

Let α, β be the roots of the equation x^2-a x-b=0 with Im(α) Im(β)...

Let

\alpha, \beta

be the roots of the equation

x^2-a x-b=0

with

\operatorname{Im}(\alpha) \lt \operatorname{Im}(\beta)

. Let

P_n=\alpha^n-\beta^n

. If

\mathrm{P}_3=-5 \sqrt{7} i, \mathrm{P}_4=-3 \sqrt{7} i, \mathrm{P}_5=11 \sqrt{7} i and

\mathrm{P}_6=45 \sqrt{7} i

, then \left|\alpha^4+\beta^4\right|

is equal to

\qquad

Correct answer

Step-by-step solution

\begin{aligned} & \alpha+\beta=\mathrm{a} \quad \alpha \beta=-\mathrm{b} \\ & \mathrm{P}_6=\mathrm{aP}_5+\mathrm{bP}_4 \\ & 45 \sqrt{7} \mathrm{i}=\mathrm{a} \times 11 \sqrt{7} \mathrm{i}+\mathrm{b}(-3 \sqrt{7}) \mathrm{i} \\ & 45=11 \mathrm{a}-3 \mathrm{~b} \end{aligned}

and

\begin{aligned} & P_5=\mathrm{aP}_4+b P_3 \\ & 11 \sqrt{7} \mathrm{i}=\mathrm{a}(-3 \sqrt{7} \mathrm{i})+\mathrm{b}(-5 \sqrt{7} \mathrm{i}) \\ & 11=-3 \mathrm{a}-5 \mathrm{~b} \\ & \mathrm{a}=3, \mathrm{~b}=-4 \\ & \left|\alpha^4+\beta^4\right|=\sqrt{\left(\alpha^4-\beta^4\right)^2+4 \alpha^4 \beta^4} \\ & =\sqrt{-63+4.4^4} \\ & =\sqrt{-63+1024}=\sqrt{961}=31 \end{aligned}

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