Question

If x=root5-2/root5+2 and y=root5+2/root5-2 then find the value of x²+y²-xy

Answer 1

$x^2+y^2-xy$ $=\frac{97}{5} -\frac{24}{\sqrt{5} }$

Step-by-step explanation:

Given,

$x= \sqrt{5}-\frac{2}{ \sqrt{5}} +2$   and   $y= \sqrt{5}+\frac{2}{ \sqrt{5}} -2$

$x-y= \sqrt{5}-\frac{2}{ \sqrt{5} }+2 -( \sqrt{5}+\frac{2}{ \sqrt{5} }-2)$

$\Leftrightarrow x-y= \sqrt{5}-\frac{2}{ \sqrt{5}} +2 - \sqrt{5}-\frac{2}{ \sqrt{5}} +2$

$\Leftrightarrow x-y= -\frac{4}{ \sqrt{5}} +4$

Therefore,

$xy=( \sqrt{5}-\frac{2}{ \sqrt{5}} +2)( \sqrt{5}+\frac{2}{ \sqrt{5}} -2)$

$\Leftrightarrow xy = (\sqrt{5})^2-(\frac{2}{\sqrt{5} } -2)^2$

$\Leftrightarrow xy =5-(\frac{4}{{5} } -\frac{8}{\sqrt{5}}+4)$

$\Leftrightarrow xy =1-\frac{4}{{5} } +\frac{8}{\sqrt{5}}$

Therefore

$x^2+y^2-xy$

$=x^2+y^2-2xy+xy$

$=(x-y)^2+xy$

$=(-\frac{4}{\sqrt{5} } +4)^2+1-\frac{4}{{5} } +\frac{8}{\sqrt{5}}$

$=\frac{16}{5} -\frac{32}{\sqrt{5}} +16+1-\frac{4}{5} +\frac{8}{\sqrt{5} }$

$=\frac{12}{5} -\frac{24}{\sqrt{5} } +17$

$=\frac{97}{5} -\frac{24}{\sqrt{5} }$