# Given x=a+1/a and y=a-1/a calculate 5(x^2+y^2)^2 ?

Mar 10, 2017

given
$x = {5}^{\frac{1}{4}} + {5}^{- \frac{1}{4}}$

=>x^2=(5^(1/4) + 5^(-1/4))^2=5^(1/2)+5^(-1/2)+2xx5^(1/4)xx5^(-1/4
and

,$y = {5}^{\frac{1}{4}} - {5}^{- \frac{1}{4}}$

,=> y^2=(5^(1/4) - 5^(-1/4))^2=5^(1/2)+5^(-1/2)-2xx5^(1/4)xx5^(-1/4

So

${x}^{2} + {y}^{2} = 2 \left({5}^{\frac{1}{2}} + {5}^{- \frac{1}{2}}\right)$

Hence

$5 {\left({x}^{2} + {y}^{2}\right)}^{2} = 5 \times 2 {\left({5}^{\frac{1}{2}} + {5}^{- \frac{1}{2}}\right)}^{2}$

$= 5 \times {2}^{2} {\left(\sqrt{5} + \frac{1}{\sqrt{5}}\right)}^{2}$

$= 5 \times {2}^{2} {\left(\frac{5 + 1}{\sqrt{5}}\right)}^{2}$

$= \cancel{5} \times {2}^{2} \times \frac{36}{\cancel{5}} = 144$

Mar 10, 2017

See below.

#### Explanation:

$x = a + \frac{1}{a}$ and $y = a - \frac{1}{a}$ so

$x y = {a}^{2} - \frac{1}{a} ^ 2$ but

${\left(x + y\right)}^{2} = {x}^{2} + {y}^{2} + 2 x y$ so

${x}^{2} + {y}^{2} = {\left(2 a\right)}^{2} - 2 \left({a}^{2} - \frac{1}{a} ^ 2\right) = 2 \left({a}^{2} + \frac{1}{a} ^ 2\right)$ and also

${\left({x}^{2} + {y}^{2}\right)}^{2} = 4 \left({a}^{4} + 2 + \frac{1}{a} ^ 4\right)$

substituting ${a}^{4} = 5$ we get at

$5 {\left({x}^{2} + {y}^{2}\right)}^{2} = 5 \cdot 4 \left(5 + 2 + \frac{1}{5}\right) = 144$