Question

If x=2+√5 , prove that x^(2)+1/x^(2)=18

Answer 1

x = 2+root5

x2 = (2+root5)2 =9 +4root5 {using identity (a+b)2}

1/x2  = (1/2+root5 )2  =1/9 +4root5 

so x2 +1/x2    = 9+4root5 + 1/9+4root5

                      =9+4root5 + 9- 4 root5 

                                          (9)2  - (4root5)2

                      =9 +4root5 +9 -4root5 

                                             1

                     9+4root5 +9 - 4root5  =18

thus x2 +1/x2 =18 proved

Answer 2

Hey mate!

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Given :

$x = 2 + \sqrt{5}$

To prove :

$x {}^{2} + \frac{1}{x {}^{2} } = 18$

Proof :

$x = 2 + \sqrt{5} \\ \\ x {}^{2} = (2 + \sqrt{5} ) {}^{2} \\ \\ x {}^{2} = 2 {}^{2} + \sqrt{5} {}^{2} + 2 \times 2 \times \sqrt{5} \\ \\ x {}^{2} = 4 + 5 + 4 \sqrt{5} \\ \\ x {}^{2} = 9 + 4 \sqrt{5}$


Now,

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

So,

$x {}^{2} + \frac{1}{x {}^{2} } = 9 + \cancel {4 \sqrt{5}} + 9 - \cancel{4 \sqrt{5}} \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 9 + 9 \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 18$


Hence, it is proved.

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Thanks for the question !

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