Question

If x = 4 - √5 and y = 4+ √5 then find :
x^4y^2 + x^2y^2

Answer 1

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$\huge\underline \mathfrak \green{Solution:-}$

Given, $x = 4 - \sqrt{5}$ and $y = 4 + \sqrt{5}$

Now,

$\red\longrightarrow\tt x + y$

$\red\longrightarrow\tt 4 - \sqrt{5} + 4 + \sqrt{5}$

$\red\longrightarrow\tt 8$

Again,

$\blue\longrightarrow\tt xy$

$\blue\longrightarrow\tt (4 - \sqrt{5} )(4 + \sqrt{5} )$

$\blue\longrightarrow\tt {4}^{2} - { (\sqrt{5} )}^{2}$

$\blue\longrightarrow\tt 16 - 5$

$\blue\longrightarrow\tt 11$

Now,

$\pink\longrightarrow\tt {x}^{4} + {y}^{4} + {x}^{2} {y}^{2}$

$\pink\longrightarrow\tt { {(x}^{2}) }^{2} + { {(y}^{2}) }^{2} + 2 {x}^{2} {y}^{2} - {x}^{2} {y}^{2}$

$\pink\longrightarrow\tt {( {x}^{2} + {y}^{2} )}^{2} - {(xy})^{2}$

$\pink\longrightarrow\tt [{ {( x + y)}^{2} - 2xy}]^{2} - {(xy)}^{2}$

$\pink\longrightarrow\tt {( {8}^{2} - 2 \times 11)}^{2} - {11}^{2}$

$\pink\longrightarrow\tt {(64 - 22)}^{2} - 121$

$\pink\longrightarrow\tt {(32)}^{2} - 121$

$\pink\longrightarrow\tt 1024 - 121$

$\pink\longrightarrow\tt 903$

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