Math, asked by Tannu8112, 5 months ago

if x= 2+√5, then find the value of x+1/x, x²+1/x² and x³+1/x³​

Answers

Answered by Anonymous
3

Solution:-

Given :-

 \sf\to \:  \: x = 2 +  \sqrt{5}

To find Value of

 \sf \to \: x +  \dfrac{1}{x}

 \sf \to \:  {x}^{2}  +  \dfrac{1}{ {x}^{2} }

And

 \sf \to \:  {x}^{3}  +  \dfrac{1}{ {x}^{3} }

Now we can write as

 \sf \to \dfrac{1}{x}  =  \dfrac{1}{2 +  \sqrt{5} }

Now rationalize the denominator

\sf \to \dfrac{1}{x}  =  \dfrac{1}{2 +  \sqrt{5} }  \times  \dfrac{2 -  \sqrt{5} }{2 -  \sqrt{5} }  =  \dfrac{2 -  \sqrt{5} }{4 - 5}

 \rm \to \:  \dfrac{1}{x} =  - 2 +  \sqrt{5}

i) Solution:-

Now put the value

 \sf \to \: x +  \dfrac{1}{x}  = 2  +  \sqrt{5}  - 2 +  \sqrt{5}

 \sf \to \: x +  \dfrac{1}{x}  = 2 \sqrt{5}

ii)Solution:-

 \sf \to \:  {x}^{2}  +  \dfrac{1}{ {x}^{2} }  =  \bigg(x +  \dfrac{1}{x}  \bigg)^{2}

We have

 \sf \to \:  \bigg(2 \sqrt{5}  \bigg)^{2}  = 4 \times 5 = 20

iii)Solution:-

 \sf \to \:  {x}^{3}  +  \dfrac{1}{ {x}^{3} }  =  \bigg(x +  \dfrac{1}{x}  \bigg)^{3}

 \rm \to \:  \bigg(2 \sqrt{5}  \bigg)^{3}  = 8 \times 5 \times  \sqrt{5}  = 40 \sqrt{5}

Answer

 \sf \to \: x +  \dfrac{1}{x}  = 2 \sqrt{5}  \:  \: , {x}^{2}  +  \dfrac{1}{ {x}^{2} }  = 20 \:  \: and \:  \:  {x}^{3}  +  \dfrac{1}{ {x}^{3} }  = 40 \sqrt{5}

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