Question

can any one solve this please

Answer 1

Hello,

The identities that we are going to use.

$( {x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$
Now,

$x = 5 - 2 \sqrt{6}$
We can find the value of 1/x

So ,

$\frac{1}{x} = \frac{1}{5 - 2 \sqrt{6} } \\ \\ rationalizing \: the \: denominator \\ \\ \frac{1}{5 - 2 \sqrt{6} } = \frac{5 + 2 \sqrt{6} }{(5 - 2 \sqrt{6})(5 + 2 \sqrt{6} )} \\ \\ = 5 + 2 \sqrt{6} \div 1 = 5 + 2 \sqrt{6}$
Now,

$x + \frac{1}{x } = 5 - 2 \sqrt{6} + 5 + 2 \sqrt{6} = 10$
Now
Squaring this

${(x + \frac{1}{x}) }^{2} = {x}^{2} + {( \frac{1}{x} )}^{2} + 2.x. \frac{1}{x} \\ {x}^{2} + \frac{1}{ {x}^{2} } = {(10)}^{2} - 2 = 100 - 2 = 98$
Therefore,
The value is equal to 98.


Hope this will be helping you ✌️