Question

X=root2-1/root2+1 and y= root2+1/root2-1. Find x^2+5xy+y^2

Answer 1

$\mathfrak{ \huge{ \bold{Answer:-} } }</p><p>\ \: \\x=\frac{\sqrt{2}-1}{\sqrt{2}+1}\\=\frac{\sqrt{2}-1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1} \\ = \frac{ {( \sqrt{2} - 1)}^{2} }{ {( \sqrt{2} )}^{2} - {(1)}^{2} } \\ = \frac{2 + 1 - 2 \sqrt{2} }{2 - 1} \\ = \frac{3 - 2 \sqrt{2} }{1} \\ = 3 - 2 \sqrt{2} \\ y = \frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1 } \\ = \frac{ \sqrt{2} + 1}{ \sqrt{2} - 1} \times \frac{ \sqrt{2} + 1 }{ \sqrt{2} + 1} \\ = \frac{ {( \sqrt{2} + 1)}^{2} }{ {( \sqrt{2}) }^{2} - {(1)}^{2} } \\ = \frac{2 + 1 + 2 \sqrt{2} }{2 - 1} \\ = \frac{3 + 2 \sqrt{2} }{1} \\ = 3 + 2 \sqrt{2} \\ {x}^{2} + 5xy + {y}^{2} \\ = {(3 - 2 \sqrt{2} )}^{2} + 5(3 - 2 \sqrt{2} )(3 + 2 \sqrt{2} ) + {(3 + 2 \sqrt{2} )}^{2} \\ = {(3)}^{2} + {(2 \sqrt{2} )}^{2} - 2(3)(2 \sqrt{2} ) + 5[ {(3)}^{2} - {(2 \sqrt{2}) }^{2} ] + {(3)}^{2} + {(2 \sqrt{2}) }^{2} + 2(3)(2 \sqrt{2} ) \\ = 9 + 8 - 12 \sqrt{2} + 5(9 - 8) + 9 + 8 + 12 \sqrt{2} \\ = 17 - 12 \sqrt{2} + 5 + 17 + 12 \sqrt{2} \\ = 17 + 17 + 5 - 12 \sqrt{2} + 12 \sqrt{2} \\ = 39 \\\mathfrak{ \bold{\large{\boxed{ {x}^{2} + 5xy + {y}^{2} = 39 }}}}$