Question 40

If $$x^{4}+\frac{1}{x^{4}}$$=47 ,then find the value of x+$$\frac{1}{x}$$

Solution

$$x^4+\frac{1}{x^4}=47$$

or, $$\left(x^2+\frac{1}{x^2}\right)^2-2.x^2.\frac{1}{x^2}=47.$$

or, $$\left(x^2+\frac{1}{x^2}\right)^2=49=7^2.$$

or, $$\left(x^2+\frac{1}{x^2}\right)=7.$$

or, $$\left(x+\frac{1}{x}\right)^2-2.x.\frac{1}{x}=7.$$

or, $$\left(x+\frac{1}{x}\right)^2=9=3^2.$$

or, $$\left(x+\frac{1}{x}\right)=3.$$

D is correct choice.

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