Verified Solution • 2025–26 Session moderate

QJEEM25Ma608c: JEE Main 2025 Mathematics - Class 11 JEEM25Ma608c Solution: Let be the roots of the equation with . Let . If and , then is…

Let α,β\alpha, \beta be the roots of the equation x2−ax−b=0x^2-\mathrm{ax}-\mathrm{b}=0 with \operatorname{Im}(\alpha)<\operatorname{Im}(\beta) . Let Pn=αn−βn\mathrm{P}_{\mathrm{n}}=\alpha^{\mathrm{n}}-\beta^{\mathrm{n}} . If P3=−57i,P4=−37i,P5=117i\mathrm{P}_3=-5 \sqrt{7} i, \mathrm{P}_4=-3 \sqrt{7} i, \mathrm{P}_5=11 \sqrt{7} i and P6=457i\mathrm{P}_6=45 \sqrt{7} i , then ∣α4+β4∣\left|\alpha^4+\beta^4\right| is equal to __________.

Step-by-Step Verified Solution

amp;α+β=aαβ=−bamp;P6=aP5+bP4amp;457i=a×117i+b(−37)iamp;45=11a−3 b.... (1)\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}

amp; and amp;amp;P5=aP4+bP3amp;117i=a(−37i)+b(−57i)amp;11=−3a−5 b.... (2)amp;a=3, b=−4amp;∣α4+β4∣=(α4−β4)2+4α4β4amp;=−63+4.44amp;=−63+1024=961=31\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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