Verified Solution • 2025–26 Session moderate QJEEM24Ma184a: JEE Main 2024 Mathematics - Class 11 JEEM24Ma184a Solution: Let be roots of . If , then is equal to ________. Let α,β\alpha, \betaα,β be roots of x2+2x−8=0x^2+\sqrt{2} x-8=0x2+2x−8=0 . If Un=αn+βn\mathrm{U}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}Un=αn+βn , then U10+2U92U8\frac{\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9}{2 \mathrm{U}_8}2U8U10+2U9 is equal to ________. Step-by-Step Verified Solution ⇒α2+2α=8\Rightarrow \alpha^2+\sqrt{2} \alpha=8⇒α2+2α=8 α+β=2,αβ=−8,amp;⇒α+2=8αamp;⇒β+2=8β\begin{aligned}\alpha+\beta=\sqrt{2}, \quad \alpha \beta=-8, & \Rightarrow \alpha+\sqrt{2}=\frac{8}{\alpha} \\& \Rightarrow \beta+\sqrt{2}=\frac{8}{\beta}\end{aligned}α+β=2,αβ=−8,amp;⇒α+2=α8amp;⇒β+2=β8 amp;U10+2U92U8=α10+β10+2α9+2β92α8+2β8amp;=α9(α+2)+β9(β+2)2(α8+β8)amp;=α9⋅(8α)+β9(8β)2(α8+β8)=82=4\begin{aligned}& \frac{U_{10}+\sqrt{2} U_9}{2 U_8}=\frac{\alpha^{10}+\beta^{10}+\sqrt{2} \alpha^9+\sqrt{2} \beta^9}{2 \alpha^8+2 \beta^8} \\& =\frac{\alpha^9(\alpha+\sqrt{2})+\beta^9(\beta+\sqrt{2})}{2\left(\alpha^8+\beta^8\right)} \\& =\frac{\alpha^9 \cdot\left(\frac{8}{\alpha}\right)+\beta^9\left(\frac{8}{\beta}\right)}{2\left(\alpha^8+\beta^8\right)}=\frac{8}{2}=4\end{aligned}amp;2U8U10+2U9=2α8+2β8α10+β10+2α9+2β9amp;=2(α8+β8)α9(α+2)+β9(β+2)amp;=2(α8+β8)α9⋅(α8)+β9(β8)=28=4 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring
