JEE MAIN - Mathematics (2025 - 23rd January Evening Shift - No. 21)
Let $\alpha, \beta$ be the roots of the equation $x^2-\mathrm{ax}-\mathrm{b}=0$ with $\operatorname{Im}(\alpha)<\operatorname{Im}(\beta)$. Let $\mathrm{P}_{\mathrm{n}}=\alpha^{\mathrm{n}}-\beta^{\mathrm{n}}$. 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 __________.
Answer
31
Explanation
$$\begin{aligned} & \alpha+\beta=\mathrm{a} \quad \alpha \beta=-\mathrm{b} \\ & \mathrm{P}_6=\mathrm{aP}_5+\mathrm{bP}_4 \\ & 45 \sqrt{7} \mathrm{i}=\mathrm{a} \times 11 \sqrt{7} \mathrm{i}+\mathrm{b}(-3 \sqrt{7}) \mathrm{i} \\ & 45=11 \mathrm{a}-3 \mathrm{~b}\quad\text{.... (1)} \end{aligned}$$
$$\begin{aligned} &\text { and }\\ &\begin{aligned} & P_5=\mathrm{aP}_4+b P_3 \\ & 11 \sqrt{7} \mathrm{i}=\mathrm{a}(-3 \sqrt{7} \mathrm{i})+\mathrm{b}(-5 \sqrt{7} \mathrm{i}) \\ & 11=-3 \mathrm{a}-5 \mathrm{~b} \quad\text{.... (2)}\\ & \mathrm{a}=3, \mathrm{~b}=-4 \\ & \left|\alpha^4+\beta^4\right|=\sqrt{\left(\alpha^4-\beta^4\right)^2+4 \alpha^4 \beta^4} \\ & =\sqrt{-63+4.4^4} \\ & =\sqrt{-63+1024}=\sqrt{961}=31 \end{aligned} \end{aligned}$$
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