If \ x^{4}+\frac{1}{x^{4}}\ = 98 and x > 1, then what is the value of x-\frac{1}{x}\ ?

Question 135

If $$\ x^{4}+\frac{1}{x^{4}}\ $$= 98 and $$x > 1$$, then what is the value of $$x-\frac{1}{x}\ ?$$

Solution

Given : $$\ x^{4}+\frac{1}{x^{4}}\ =98$$

=> $$(x^2+\frac{1}{x^2})^2-2(x^2)(\frac{1}{x^2})=98$$

=> $$(x^2+\frac{1}{x^2})^2=98+2=100$$

=> $$x^2+\frac{1}{x^2}=\sqrt{100}=10$$

=> $$(x-\frac{1}{x})^2+2(x)(\frac{1}{x})=10$$

=> $$(x-\frac{1}{x})^2=10-2=8$$

=> $$x-\frac{1}{x}=\sqrt8=2\sqrt2$$

=> Ans - (B)

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