$\begin{aligned}
& \alpha+\beta=\mathrm{a} \quad \alpha \beta=-\mathrm{b} \
& \mathrm{P}_6=\mathrm{aP}_5+\mathrm{bP}_4 \
& 45 \sqrt{7} \mathrm{i}=\mathrm{a} \times 11 \sqrt{7} \mathrm{i}+\mathrm{b}(-3 \sqrt{7}) \mathrm{i} \
& 45=11 \mathrm{a}-3 \mathrm{b}
\end{aligned}and\begin{aligned}
& P_5=\mathrm{aP}_4+b P_3 \
& 11 \sqrt{7} \mathrm{i}=\mathrm{a}(-3 \sqrt{7} \mathrm{i})+\mathrm{b}(-5 \sqrt{7} \mathrm{i}) \
& 11=-3 \mathrm{a}-5 \mathrm{b} \
& \mathrm{a}=3, \mathrm{~b}=-4 \
& \left|\alpha^4+\beta^4\right|=\sqrt{\left(\alpha^4-\beta^4\right)^2+4 \alpha^4 \beta^4} \
& =\sqrt{-63+4.4^4} \
& =\sqrt{-63+1024}=\sqrt{961}=31
\end{aligned}$
Let α, β be the roots of the equation x^2-a x-b=0 with Im(α) lt 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,…
JEE Main 2025 — Mathematics Algebra
Let α,β be the roots of the equation x2−ax−b=0 with Im(α)<Im(β). Let Pn=αn−βn. If P3=−57i,P4=−37i,P5=117i and P6=457i, then α4+β4 is equal to .
Held on 23 Jan 2025 · Verified 6 Jul 2026.
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