Question

If x= 2+ √3 then find out
(I) x^4+1/x^4​

Answer 1

Answer:

194

I really hope it helps.

Answer 2

Step-by-step explanation:

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

$\\ \sf \underline{On \: Rationalising \: we \: will \: get:-}$

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

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

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

$\sf Squaring \: x+ \dfrac{1}{x} \: will \: give :-$

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

$\sf We \: know \: that,$

$\sf (a+b)^2=a^2+b^2+2ab$

$\sf Using \: this \: identity,$

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

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

$\sf \implies (x+ \dfrac{1}{x})^2=( (2)^2+(\sqrt{3})^2+2 \times 2 \times \sqrt{3})+( {(2)}^{2}+ {(\sqrt{3})}^{2}-2 \times 2 \times \sqrt{3})+2 \\ \\ \sf {(x+ \dfrac{1}{x})}^{2}=(4+3+4 \sqrt{3})+(4+3-4 \sqrt{3})+2 \\ \\ \sf \implies {(x+ \dfrac{1}{x})}^{2}=7+7+2 \\ \\ \sf \implies (x+ \dfrac{1}{x})^2=16$

$\sf \implies 16= x^2 + \dfrac{1}{x^2} + 2 \\ \\ \sf \implies 16-2=x^2+ \dfrac{1}{x^2} \\ \\ \sf \implies x^2+ \dfrac{1}{x^2}=14$

$\sf Squaring \: x^2 + \dfrac{1}{x^2} \: will \: give :-$

$\sf (x^2+ \dfrac{1}{x^2})^2$

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

$\longmapsto \sf x^4+ \dfrac{1}{x^4}=(14)^2-2 \\ \\ \sf \longmapsto x^4+ \dfrac{1}{x^4}=196-2 \\ \\ \sf \longmapsto x^4 + \dfrac{1}{x^4}=194$

Hope it helps.

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