Math, asked by kajaljyala12345, 6 months ago

If x=√6+√5 .then find x²+ 1/x² -2.​

Answers

Answered by pk695743mayank
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

Answered by mathdude500
0

\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} }

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