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Question

If $$x + \dfrac{1}{x} = -2 $$, find the value of $$(x^2 + \dfrac{1}{x^2})\times$$$$(x^4 + \dfrac{1}{x^4})$$


A
2
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B
1
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C
4
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D
1
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Solution

The correct option is C $$4$$
Given $$x+\dfrac 1x=-2$$

squaring on both sides

$$(x+\dfrac 1x)^2=(-2)^2$$

$$\implies x^2+\dfrac{1}{x^2}+2=4$$

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

squaring on both sides

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

$$\implies x^4+\dfrac{1}{x^4}+2=4$$

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

product of (1) and (2) is 

$$(x^2+\dfrac{1}{x^2})(x^4+\dfrac{1}{x^4})=2*2=4$$ 

Mathematics

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