Question

if x=5+2√6 and y=1/x find the value of x²+y²​

Answer 1

Answer:

see the pic above

Step-by-step explanation:

hope it helps ✌️

Answer 2

Given

$\tt \to \: x = 5 + 2 \sqrt{6}$

$\tt \to \: y = \dfrac{1}{x}$

To Find the value

$\tt \to \: {x}^{2} + {y}^{2}$

According to question

$\tt \to \: {x}^{2} + \dfrac{1}{ {x}^{2} }$

if

$\tt \to \: x = 5 + 2 \sqrt{6}$

Then

$\tt \to \: \dfrac{1}{x}= \dfrac{1}{5 + 2 \sqrt{6} }$

Rationalize The Denominator

$\tt \to \: \dfrac{1}{x}= \dfrac{1}{5 + 2 \sqrt{6} } \times \dfrac{5 - 2 \sqrt{6} }{5 - 2 \sqrt{6} }$

$\tt \to \: \dfrac{1}{x}= \dfrac{5 - 2 \sqrt{6} }{(5 ) {}^{2} - (2 \sqrt{6}) {}^{2} }$

$\tt \to \: \dfrac{1}{x}= \dfrac{5 - 2 \sqrt{6} }{25 - 24 } = 5 - 2 \sqrt{6}$

Now we have to find

$\tt \to \: {x}^{2} + \dfrac{1}{ {x}^{2} }$

we can write as

$\tt \to \bigg(x + \dfrac{1}{x} \bigg) ^{2} = {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x}$

$\tt \to \bigg(x + \dfrac{1}{x} \bigg) ^{2} = {x}^{2} + \dfrac{1}{ {x}^{2} } + 2$

$\tt \to \bigg(x + \dfrac{1}{x} \bigg) ^{2} - 2 = {x}^{2} + \dfrac{1}{ {x}^{2} }$

Put the value

$\tt \to \bigg(5 + 2 \sqrt{6} + 5 - 2 \sqrt{6} \bigg)^{2} - 2 = {x}^{2} + \dfrac{1}{ {x}^{2} }$

$\tt\to\:{x}^{2}+\dfrac{1}{{x}^{2}}=(10)^{2}-2$

$\tt\to\:{x}^{2}+\dfrac{1}{{x}^{2}} =100-2$

$\tt\to\:{x}^{2}+\dfrac{1}{{x}^{2}}=98$

Answer

$\tt\to\:{x}^{2}+\dfrac{1}{{x}^{2}}=98$