Question

find the value of x square +1/x square when x=4 +root 15​

Answer 1

Given:

• x = 4 + √15

To find:

• x² + 1/x²

Method:

First, let us find out 1/x

$\sf{\dfrac{1}{x} = \dfrac{1}{4 + \sqrt{15} } = \dfrac{1}{4 + \sqrt{15} } \times \dfrac{4 - \sqrt{15} }{4 - \sqrt{15} } = \dfrac {4 - \sqrt{15}}{16 - 15} }\\\\\\\implies \bf{\dfrac{1}{x} = 4 - \sqrt{15} }$

x + 1/x

= 4 + √15 + 4 - √15

x + 1/x = 8

$(\sf{x + \dfrac{1}{x})^{2} = x^{2} + \dfrac{1}{x^{2} }+ 2}\\\\\\\implies \sf{(8)^{2} = x^{2} + \dfrac{1}{x^{2} } + 2 }\\\\\\\implies \sf{64 = x^{2} + \dfrac{1}{x^{2} } + 2}\\\\\\\implies \sf{x^{2} + \dfrac{1}{x^{2}}= 64 - 2}\\\\\\\implies \boxed{\bf{x^{2} + \dfrac{1}{x^{2}}= 62} }$

Answer 2

Solution

Given ,

• x = 4+√15

To find ,

• x^2 + 1/x^2

So,

• x = 4+√15
• 1/x = 1/4+√15

By rationalisation ;

$= > \frac{1}{4 + \sqrt{15} } \times \frac{4 - \sqrt{15} }{4 - \sqrt{15} }$

By [ (a+b)(a-b) a^2 - b^2 ]

$= > \frac{4 - \sqrt{15} }{( {4}^{2}) - ( \sqrt{15 )^{2} } }$

$= > \frac{4 - \sqrt{15} }{16 - 15}$

$\bold{ = > 4 - \sqrt{15} = \frac{1}{x} }$

Now ,

• x + 1/x = 4+√15+4-√15

Solving this ;

$= > x + \frac{1}{x} = 4 + \cancel{ \sqrt{15} } + 4 - \cancel{ \sqrt{15} }$

$= > {x}^{2} + \frac{1}{ {x}^{2} } = {8}^{2} - 2$

$= > {x}^{2} + \frac{1}{ {x}^{2} } = 64 - 2$

$\bold{ = > {x}^{2} + \frac{1}{ {x}^{2} } = 62}$

The value is x^2 + 1/x^2 = 62 .