JEE Main 2025 3 April Morning Shift
Exam JEE Main
Year 2025
Date 3 April
Shift Morning Shift
Subject Mathematics
Chapter Quadratic Equation JEE Main 2025 · 3 April · Morning Shift · Mathematics · Quadratic Equation
Let α and β be the roots of x^2+ 3 x -16=0 , and γ and be the roots...
Let
α \alpha α and
β \beta β be the roots of
x 2 + 3 x − 16 = 0 x^2+\sqrt{3 x}-16=0 x 2 + 3 x − 16 = 0 , and
γ \gamma γ and
δ \delta δ be the roots of
x 2 + 3 x − 1 = 0 x^2+3 x-1=0 x 2 + 3 x − 1 = 0 . If
P n = α n + β n P_n=\alpha^n+\beta^n P n = α n + β n and
Q n = γ n + δ n Q_n=\gamma^n+\delta^n Q n = γ n + δ n , then
P 25 + 3 P 24 2 P 23 + Q 25 − Q 23 Q 24 \frac{\mathrm{P}_{25}+\sqrt{3 \mathrm{P}_{24}}}{2 \mathrm{P}_{23}}+\frac{\mathrm{Q}_{25}-\mathrm{Q}_{23}}{\mathrm{Q}_{24}} 2 P 23 P 25 + 3 P 24 + Q 24 Q 25 − Q 23 is equal to
Step-by-step solution
x 2 + 3 x − 16 = 0 < β P n + 3 P n − 1 − 16 P n − 2 = 0 P 25 + 3 P 24 − 16 P 23 = 0 ∴ P 25 + 3 P 24 2 P 23 = 8 \begin{aligned} & x^2+\sqrt{3} \mathrm{x}-16=0 \lt \beta \\ & \mathrm{P}_{\mathrm{n}}+\sqrt{3} \mathrm{P}_{\mathrm{n}-1}-16 \mathrm{P}_{\mathrm{n}-2}=0 \\ & \mathrm{P}_{25}+\sqrt{3} \mathrm{P}_{24}-16 \mathrm{P}_{23}=0 \\ & \therefore \frac{\mathrm{P}_{25}+\sqrt{3} \mathrm{P}_{24}}{2 \mathrm{P}_{23}}=8\end{aligned} x 2 + 3 x − 16 = 0 < β P n + 3 P n − 1 − 16 P n − 2 = 0 P 25 + 3 P 24 − 16 P 23 = 0 ∴ 2 P 23 P 25 + 3 P 24 = 8 Similarly
x 2 + 3 x − 1 = 0 < ∑ δ γ Q n = γ n + δ n x^2+3 x-1=0 \lt \sum_\delta^\gamma \quad Q_n=\gamma^n+\delta^n x 2 + 3 x − 1 = 0 < ∑ δ γ Q n = γ n + δ n < b r / > Q 25 − Q 23 = γ 25 + δ 25 − γ 23 − δ 23 = γ 23 ( γ 2 − 1 ) + δ 23 ( δ 2 − 1 ) = γ 23 ( − 3 γ ) + δ 23 ( − 3 γ ) = − 3 [ γ 24 + δ 24 ] = − 3 Q 24 ∴ Q 25 − Q 23 Q 24 = − 3 < b r / > \begin{aligned}<br/>& \mathrm{Q}_{25}-\mathrm{Q}_{23}=\gamma^{25}+\delta^{25}-\gamma^{23}-\delta^{23} \\
&=\gamma^{23}\left(\gamma^2-1\right)+\delta^{23}\left(\delta^2-1\right) \\
&=\gamma^{23}(-3 \gamma)+\delta^{23}(-3 \gamma) \\
&=-3\left[\gamma^{24}+\delta^{24}\right] \\
&=-3 \mathrm{Q}_{24} \\
& \therefore \frac{\mathrm{Q}_{25}-\mathrm{Q}_{23}}{\mathrm{Q}_{24}}=-3<br/>\end{aligned} < b r / > Q 25 − Q 23 = γ 25 + δ 25 − γ 23 − δ 23 = γ 23 ( γ 2 − 1 ) + δ 23 ( δ 2 − 1 ) = γ 23 ( − 3 γ ) + δ 23 ( − 3 γ ) = − 3 [ γ 24 + δ 24 ] = − 3 Q 24 ∴ Q 24 Q 25 − Q 23 = − 3 < b r / > P 25 + 3 P 24 2 P 23 + Q 25 − Q 23 Q 24 = 8 − 3 = 5 \frac{\mathrm{P}_{25}+\sqrt{3} \mathrm{P}_{24}}{2 \mathrm{P}_{23}}+\frac{\mathrm{Q}_{25}-\mathrm{Q}_{23}}{\mathrm{Q}_{24}}=8-3=5 2 P 23 P 25 + 3 P 24 + Q 24 Q 25 − Q 23 = 8 − 3 = 5
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