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4-2.Quadratic Equations and Inequations
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ધારો કે $\alpha, \beta$ એ $x^2+\sqrt{2} x-8=0$ નાં બીજ છે. જો $\mathrm{U}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}$, તો $\frac{\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9}{2 \mathrm{U}_8}=$...........
A
$5$
B
$9$
C
$44$
D
$4$
(JEE MAIN-2024)
Solution
$ \frac{\alpha^{10}+\beta^{10}+\sqrt{2}\left(\alpha^9+\beta^9\right)}{2\left(\alpha^8+\beta^8\right)} $
$ \frac{\alpha^8\left(\alpha^2+\sqrt{2} \alpha\right)+\beta^8\left(\beta^2+\sqrt{2} \beta\right)}{2\left(\alpha^8+\beta^8\right)} $
$ \frac{8 \alpha^8+8 \beta^8}{2\left(\alpha^8+\beta^8\right)}=4$
Standard 11
Mathematics