Question

If x=√6+√5 .then find x²+ 1/x² -2.​

Answer 1

Answer:

1.2 is the answer

Step-by-step explanation:

x=root 6 +root 5--------1

put x =1 equation

x^2+1/x^2-2

(6+5)+1/(6+5)-2

11+1/11-2

12/10

1.2

Answer 2

$\begin{gathered}\begin{gathered}\bf \: Given - \begin{cases} &\sf{x = \sqrt{6} + \sqrt{5} } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\:find - \begin{cases} &\sf{ {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 }  \end{cases}\end{gathered}\end{gathered}$

Concept Used :-

Method of Rationalization :-

• Rationalization is the process of eliminating a radical or imaginary number from the denominator or numerator of an algebraic fraction. That is, remove the radicals in a fraction so that the denominator or numerator only contains a rational number.

• The denominator of the above fraction has a binomial radical i.e., is the sum of two terms, one of which is an irrational number.

• Multiply the numerator and denominator of the fraction with the conjugate of the radical.

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$

$\boxed{ \bf \: {(x + y)}^{2}+{(x - y)}^{2} = 2( {x}^{2}+{y}^{2})}$

$\large\underline{\bold{Solution-}}$

Given that

$\rm :\longmapsto\:x \: = \: \sqrt{6} + \sqrt{5}$

So,

$\rm :\longmapsto\:\dfrac{1}{x} = \dfrac{1}{ \sqrt{6} + \sqrt{5} }$

• On rationalizing the denominator, we get

$\rm :\longmapsto\:\dfrac{1}{x} = \dfrac{1}{ \sqrt{6} + \sqrt{5} } \times \dfrac{ \sqrt{6} - \sqrt{5} }{ \sqrt{6} - \sqrt{5} }$

$\rm :\longmapsto\:\dfrac{1}{x} = \dfrac{ \sqrt{6} - \sqrt{5} }{ {( \sqrt{6} )}^{2} - {( \sqrt{5}) }^{2} }$

$\rm :\longmapsto\:\dfrac{1}{x} = \dfrac{ \sqrt{6} - \sqrt{5} }{6 - 5}$

$\rm :\longmapsto\:\dfrac{1}{x} = \sqrt{6} - \sqrt{5}$

Now,

• Consider,

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2$

$\rm :\longmapsto\: {\bigg( \sqrt{6} + \sqrt{5} \bigg) }^{2} + {\bigg( \sqrt{6} - \sqrt{5} \bigg) }^{2} - 2$

$\rm :\longmapsto\:2\bigg( {( \sqrt{6}) }^{2} + {( \sqrt{5}) }^{2} \bigg) - 2$

$\rm :\longmapsto\:2(6 + 5) - 2$

$\rm :\longmapsto\:22 - 2$

$\rm :\longmapsto\:20$

Alternative method

We have to find the value of

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2$

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 \times x \times \dfrac{1}{x}$

$\rm :\longmapsto\: {\bigg(x - \dfrac{1}{x} \bigg) }^{2}$

$\rm :\longmapsto\: {\bigg( \sqrt{6} + \sqrt{5} - \sqrt{6} + \sqrt{5} \bigg) }^{2}$

$\rm :\longmapsto\: {\bigg( 2 \sqrt{5} \bigg) }^{2}$

$\rm :\longmapsto\:4 \times 5$

$\rm :\longmapsto\:20$

Useful Identities :-

$\boxed{ \bf \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy}$

$\boxed{ \bf \: {(x + y)}^{2} = {x}^{2}+{y}^{2} + 2xy}$

$\boxed{ \bf \: {(x - y)}^{2} = {x}^{2}+{y}^{2} - 2xy}$

$\boxed{ \bf \: (x + y)(x - y)= {x}^{2} - {y}^{2} }$