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 \ \ \ \ \ .

Mathematics · Quadratic Equations

JEE Main 2025 — 23 January, Evening Shift — Question 22

Let $\alpha, \beta$ be the roots of the equation $x^{2}-a x-b=0$ with $\operatorname{Im}(\alpha)<\operatorname{Im}(\beta)$ . Let

$P_{n}=\alpha^{n}-\beta^{n}$ . If $P_{3}=-5 \sqrt{7} i, \quad P_{4}=-3 \sqrt{7} i, \quad P_{5}=11 \sqrt{7} i \quad$ and $P_{6}=45 \sqrt{7} i$ , then

$\left|\alpha^{4}+\beta^{4}\right|$ is equal to $\_\_\_\_\_$ .

Numerical answer — enter this value.

$\quad \alpha+\beta=\mathrm{a} \quad \alpha \beta=-\mathrm{b}$

$\mathrm{P}_{6}=\mathrm{aP}_{5}+\mathrm{bP}_{4}$

$45 \sqrt{7} \mathrm{i}=\mathrm{a} \times 11 \sqrt{7} \mathrm{i}+\mathrm{b}(-3 \sqrt{7}) \mathrm{i}$

$45=11 \mathrm{a}-3 \mathrm{~b}$

and

$\mathrm{P}_{5}=\mathrm{aP}_{4}+\mathrm{bP}_{3}$

$11 \sqrt{7} \mathrm{i}=\mathrm{a}(-3 \sqrt{7} \mathrm{i})+\mathrm{b}(-5 \sqrt{7} \mathrm{i})$

$11=-3 \mathrm{a}-5 \mathrm{~b}$

$\mathrm{a}=3, \mathrm{~b}=-4$

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

$=\sqrt{-63+4.4^{4}}$

$=\sqrt{-63+1024}=\sqrt{961}=31$

Answer key and solution verified before publishing.

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