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
B
1
C
4
D
1

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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