Question

If x= 2-√3 find x power 4+1 divisible by x power 4​

Answer 1

★Given:-

$\sf x = 2 - \sqrt{3}$

★To find:-

$\sf x^4 + \dfrac{1}{x^4}$

★Solution:-

$\sf x = 2 - \sqrt{3}$

then,

$\sf \dfrac{1}{x} = \dfrac{1}{2 - \sqrt{3}}$

$\sf Rationalising \: the \: denominator, \: we \: will \: get,$

$\sf \dfrac{1}{x} = \dfrac{1}{2 - \sqrt{3}} \times \dfrac{2 + \sqrt{3}}{2 + \sqrt{3}} = \dfrac{2+ \sqrt{3}}{(2)^2 - ( \sqrt{3} )^2}$

$\sf \implies \dfrac{2 + \sqrt{3}}{4-3} = \dfrac{2+ \sqrt{3}}{1} = 2+ \sqrt{3}$

$\sf \dfrac{1}{x} = 2+ \sqrt{3}$

_______________________________

Now we will find $\sf x^4 + \dfrac{1}{x^4}$

$\implies \sf x^4 + \dfrac{1}{x^4} = \{ (x)^2 \}^2 + \Bigg \{ \Bigg ( \dfrac{1}{x} \Bigg )^2 \Bigg \} ^2$

$\implies \sf x^4 + \dfrac{1}{x^4} = \{ (2- \sqrt{3})^2 \}^2 + \{ ( 2 + \sqrt{3} )^2 \}^2$

$\implies \sf x^4 + \dfrac{1}{x^4} = \{ (2)^2 + (\sqrt{3})^2 - 2 \times 2 \times \sqrt{3} \}^2 + \{ (2)^2 + ( \sqrt{3} )^2 + 2 \times 2 \times \sqrt{3} \}^2$

$\implies \sf x^4 + \dfrac{1}{x^4} = (4 + 3 -4 \sqrt{3})^2 + ( 4 +3 +4 \sqrt{3} )^2$

$\implies \sf x^4 + \dfrac{1}{x^4} = (7-4 \sqrt{3})^2 + (7 +4 \sqrt{3} )^2$

$\implies \sf x^4 + \dfrac{1}{x^4} = (7)^2 + (4 \sqrt{3})^2 - 2 \times 7 \times 4 \sqrt{3} + (7)^2 + (4 \sqrt{3} )^2 + 2 \times 7 \times 4 \sqrt{3}$

$\implies \sf x^4 + \dfrac{1}{x^4} = 49 + 48 \: \cancel{- 56 \sqrt{3}} + 49 + 48 \: \cancel{+ 56 \sqrt{3}}$

$\implies \sf x^4 + \dfrac{1}{x^4} = 49 + 48 + 49 + 48$

$\implies \sf x^4 + \dfrac{1}{x^4} = 194$

$\boxed{\bold{x^4 + \dfrac{1}{x^4} = 194}}$

Hope it helps