Question 115

If for non-zero x, $$x^2-4x-1=0$$ the value of $$x^2+\frac{1}{x^2}$$ is:

Solution

Given : $$x^2-4x-1=0$$

=> $$x^2-1=4x$$

=> $$\frac{x^2-1}{x}=4$$

=> $$x-\frac{1}{x}=4$$

Squaring both sides, we get :

=> $$(x-\frac{1}{x})^2=(4)^2$$

=> $$x^2+\frac{1}{x^2}-2(x)(\frac{1}{x})=16$$

=> $$x^2+\frac{1}{x^2}=16+2=18$$

=> Ans - (D)

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