Question

If x is equal to root 5 + 2 then prove that x square + 1 divided by x square is equal to 18

Answer 1

Solution:

Given that ;

x = √5 + 2

Now, find the value of 1/x.

$\sf \implies \frac{1}{x} = \frac{1}{ \sqrt{5} + 2} \\ \\ \sf \implies \frac{1}{x} = \frac{1}{ \sqrt{5} + 2 } \times \frac{ \sqrt{5} - 2 }{ \sqrt{5} - 2 } \\ \\ \sf \implies \frac{1}{x} = \frac{ \sqrt{5} - 2 }{ {( \sqrt{5}) }^{2} - {(2)}^{2} } \\ \\ \sf \implies \frac{1}{ x } = \frac{ \sqrt{5} - 2}{5 - 4} \\ \\ \sf \implies \frac{1}{x} = \sqrt{5} - 2$

So, find x + (1/x).

$\sf \implies x + \frac{1}{x} = \sqrt{5} + 2 + \sqrt{5} - 2 \\ \\ \sf \implies x + \frac{1}{x} = 2 \sqrt{5}$

On squaring both sides ;

$\sf \implies (x + \frac{1}{x} ) {}^{2} = (2 \sqrt{5} ) {}^{2} \\ \\ \sf \implies {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 20 \\ \\ \sf \implies {x}^{2} + \frac{1}{ {x}^{2} } = 20 - 2 \\ \\ \boxed{ \red{ \bf \therefore \: {x}^{2} + \frac{1}{ {x}^{2} } = 18}}$

Hence, it is proved.