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

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Answer 1 · 2017-03-10T11:28:35

given
$x=5^(1/4) + 5^(-1/4)$

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

, $y=5^(1/4) - 5^(-1/4)$

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

So

$x^2+y^2=2(5^(1/2)+5^(-1/2))$

Hence

$5(x^2+y^2)^2=5xx2(5^(1/2)+5^(-1/2))^2$

$=5xx2^2(sqrt5+1/sqrt5)^2$

$=5xx2^2((5+1)/sqrt5)^2$

$=cancel5xx2^2xx36/cancel5=144$

Answer 2 · 2017-03-10T12:01:33

See below.

$x=a+1/a$ and $y=a-1/a$ so

$xy=a^2-1/a^2$ but

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

$x^2+y^2=(2a)^2-2(a^2-1/a^2) = 2(a^2+1/a^2)$ and also

$(x^2+y^2)^2=4(a^4+2+1/a^4)$

substituting $a^4=5$ we get at

$5(x^2+y^2)^2=5 cdot 4(5+2+1/5)=144$

Given x=a+1/ax=a+1/a and y=a-1/ay=a-1/a calculate 5(x^2+y^2)^25(x^2+y^2)^2 ?