Question

x square- 5x- 1=0 find xpower 11 +xpower 9 - 727x
power 6- 727 xpower 4

Answer 1

Given:

x square- 5x- 1=0

To find:

xpower 11 +xpower 9 - 727x  power 6- 727 xpower 4

Solution:

$x^2-5x-1=0\\\\\\x_{1,\:2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\cdot \:1\left(-1\right)}}{2\cdot \:1}:\quad \dfrac{5+\sqrt{29}}{2}\\\\\\x=\dfrac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:1\left(-1\right)}}{2\cdot \:1}:\quad \dfrac{5-\sqrt{29}}{2}\\\\\\\\x=\dfrac{5+\sqrt{29}}{2},\:x=\dfrac{5-\sqrt{29}}{2}$

Let us consider the positive value of x, so we have,

xpower 11 +xpower 9 - 727x power 6- 727 xpower 4

$\left(\dfrac{5+\sqrt{29}}{2}\right)^{11}+\left(\dfrac{5+\sqrt{29}}{2}\right)^9-727\left(\dfrac{5+\sqrt{29}}{2}\right)^6-727\left(\dfrac{5+\sqrt{29}}{2}\right)^4\\\\\\=\dfrac{\left(5+\sqrt{29}\right)^{11}}{2^{11}}+\dfrac{\left(5+\sqrt{29}\right)^9}{2^9}-\dfrac{727\left(5+\sqrt{29}\right)^6}{64}-\dfrac{727\left(5+\sqrt{29}\right)^4}{16}$

$=\dfrac{\left(5+\sqrt{29}\right)^{11}}{2048}+\dfrac{\left(5+\sqrt{29}\right)^9}{512}-\dfrac{\left(5+\sqrt{29}\right)^6\cdot \:727}{64}-\dfrac{\left(5+\sqrt{29}\right)^4\cdot \:727}{16}$

$=\dfrac{\left(5+\sqrt{29}\right)^{11}}{2048}+\dfrac{\left(5+\sqrt{29}\right)^9\cdot \:4}{2048}-\dfrac{23264\left(5+\sqrt{29}\right)^6}{2048}-\dfrac{93056\left(5+\sqrt{29}\right)^4}{2048}$

$=\dfrac{\left(5+\sqrt{29}\right)^{11}+\left(5+\sqrt{29}\right)^9\cdot \:4-23264\left(5+\sqrt{29}\right)^6-93056\left(5+\sqrt{29}\right)^4}{2048}$

$=\dfrac{63449899008+11782350848\sqrt{29}}{2048}$