If alpha, beta are roots of the equation x^{2 | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$x^{2}+5 \sqrt{2} x+10=0$

$\& p_{n}=\alpha^{n}-\beta^{n} 	ext { (Given) }$

$	ext { Now } \frac{P_{17} P_{20}+5 \sqrt{2} p_{11} P_{19}}{P_{

Vedclass · Accepted Answer

$1$

Vedclass · Answer

$x^{2}+5 \sqrt{2} x+10=0$

$\& p_{n}=\alpha^{n}-\beta^{n} 	ext { (Given) }$

$	ext { Now } \frac{P_{17} P_{20}+5 \sqrt{2} p_{11} P_{19}}{P_{18} P_{19}+5 \sqrt{2} P_{18}^{2}}=\frac{P_{17}\left(P_{20} 5 \sqrt{2} P_{19}ight)}{P_{18}\left(P_{19}+5 \sqrt{2 P}_{18}ight)}$

$\frac{P_{17}\left(\alpha^{20}-\beta^{20}+5 \sqrt{2}\left(\alpha^{19}-\beta^{19}ight)ight)}{P_{18}\left(\alpha^{19}-\beta^{19}+5 \sqrt{2}\left(\alpha^{18}-\beta^{18}ight)ight)}$

$\frac{P_{17}\left(\alpha^{19}(\alpha+5 \sqrt{2})-\beta^{19}(\beta+5 \sqrt{2})ight)}{P_{18}\left(\alpha^{18}(\alpha+5 \sqrt{2})-\beta^{18}(\beta+5 \sqrt{2})ight)}$

$	ext { Since } \alpha+5 \sqrt{2}=-10 / \alpha \ldots \ldots \ldots \ldots . .(1)$

$\&\, \beta+5 \sqrt{2}=-10 / \beta \ldots \ldots \ldots \ldots(2)$

Now put there values in above expression

$=-\frac{10 P_{1} P_{18}}{-10 P_{18} P_{17}}=1$

If $\alpha, \beta$ are roots of the equation $x^{2}+5 \sqrt{2} x+10=0, \alpha\,>\,\beta$ and $P_{n}=\alpha^{n}-\beta^{n}$ for each positive integer $\mathrm{n}$, then the value of $\left(\frac{P_{17} P_{20}+5 \sqrt{2} P_{11} P_{19}}{P_{18} P_{19}+5 \sqrt{2} P_{18}^{2}}\right)$ is equal to $....$

Similar Questions

If $x$ is real, then the value of $\frac{{{x^2} + 34x - 71}}{{{x^2} + 2x - 7}}$ does not lie between

If the set of all values of $a$, for which the equation $5 x ^3-15 x - a =0$ has three distinct real roots, is the interval $(\alpha, \beta)$, then $\beta-2 \alpha$ is equal to _____

If $x$ is real , the maximum value of $\frac{{3{x^2} + 9x + 17}}{{3{x^2} + 9x + 7}}$ is

The number of non-negative integer solutions of the equations $6 x+4 y+z=200$ and $x+y+z=100$ is

Let $\alpha$ and $\beta$ be the roots of the equation $5 x^{2}+6 x-2=0 .$ If $S_{n}=\alpha^{n}+\beta^{n}, n=1,2,3 \ldots$ then :