Question

If x=root5+2,then prove that x sq. +1/x sq.=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.

Answer 2

Answer:

$\large (x^2+\dfrac{1}{x^2})=18 \ (Proved)$

Step-by-step explanation:

Given :

$\large x=\sqrt5+2$

We have to prove

$\large x^2+(\dfrac{1}{x})^2=18$

First take a reciprocal in both of given

$\large x=\sqrt5+2\\\\\\\large \dfrac{1}{x} =\dfrac{1}{\sqrt5+2}\\\\\\\large \text{Now rationalize the denominator}\\\\\\\large \dfrac{1}{x} =\dfrac{1}{\sqrt5+2}\times\dfrac{\sqrt5-2}{\sqrt5-2}\\\\\\\large \text{Using identity (a+b)(a-b)=a^2-b^2 \ here \ we \ get }\\\\\\\large \dfrac{1}{x}=\dfrac{\sqrt5-2}{5-4}\\\\\\\large \dfrac{1}{x}=\sqrt5-2$

$\large \text{Now subtract (x-\dfrac{1}{x}) \ we \ get }$

$\large x-\dfrac{1}{x}=\sqrt5+2-(\sqrt5-2)\\\\\\\large x-\dfrac{1}{x}=4\\\\\\\large \text{Now squaring on both side we get}\\\\\\\large (x-\dfrac{1}{x})2=(4)^2\\\\\\\large (x^2+\dfrac{1}{x^2}-2)=16\\\\\\\large (x^2+\dfrac{1}{x^2})=16+2\\\\\\\large (x^2+\dfrac{1}{x^2})=18$

Hence proved.