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Question

If $$\displaystyle \left (x + \dfrac{1}{x} \right ) = 4$$, find the value of $$\displaystyle \left (x^4 + \dfrac{1}{x^4} \right )$$


A
120
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B
194
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C
160
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D
142
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Solution

The correct option is B $$194$$
$$x + \dfrac{1}{x} = 4$$
Squaring both sides,
$$=(x + \dfrac{1}{x})^2 = 4^2$$
$$=x^2 + (\dfrac{1}{x})^2+2(x)(\dfrac{1}{x})= 16$$
$$=x^2 + \dfrac{1}{x^2}+2=16$$
$$=x^2 + \dfrac{1}{x^2}=14$$
Again Squaring both sides,we get,
$$=(x^2 + \dfrac{1}{x^2})^2=14^2$$
$$=(x^2)^2 +( \dfrac{1}{x^2})^2+2(x^2)( \dfrac{1}{x^2})=196$$
$$=x^4+\dfrac{1}{x^4}=196-2$$
$$=x^4+\dfrac{1}{x^4}=194$$

Mathematics

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