Question

if x=9+4√5 and y=1/x, find x+ y, x- y, x²- y² and (x+y)²




give me answer plz fast it's urgent ​

Answer 1

$\frac{1}{x} = \frac{1}{9 + 4 \sqrt{5} } \\ \\ = > \frac{1}{9 + 4 \sqrt{5} } \times \frac{9 - 4 \sqrt{5} }{9 - 4 \sqrt{5} }$

By rationalising the denominator,

$= > \frac{1 \times (9 - 4 \sqrt{5} )}{ {9}^{2} - {(4 \sqrt{5} )}^{2} } \\ \\ = > \frac{9 - 4 \sqrt{5} }{81 - 80} = 9 - 4 \sqrt{5} = y$

Therefore, By substituting the values,

$x + y =( 9 + 4 \sqrt{5}) + (9 - 4 \sqrt{5} ) = 18$

$x - y = (9 + 4 \sqrt{5} ) - (9 - 4 \sqrt{5} ) \\ \\ = > 9 + 4 \sqrt{5} - 9 + 4 \sqrt{5} = 8 \sqrt{5}$

${x}^{2} - {y}^{2} = ({9 + 4 \sqrt{5}) }^{2} - {(9 - 4 \sqrt{5}) }^{2} \\ \\ = > (81 + 80) - (81 - 80) \\ \\ = > 81 + 80 - 81 + 80 = 160$

$(x + y)^{2} = {18}^{2} = 324 \: \: \: \: \: \: \: (we \: already \: found \: (x + y) = 18$

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