Question

If $$x^2+\frac{1}{x^2}=18,$$  find the values of $$x+\frac{1}{x}\text{ and }x-\frac{1}{x}.$$

Answer 1

Let us consider the following expression: $$x+\frac{1}{x}$$

Squaring the above expression, we get:

$${(x+\frac{1}{x})}^2=x^2+2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=x^2-2+\frac{1}{x^2}[(a+b{)}^2=a^2+b^2+2ab]$$

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

$$\Rightarrow {(x+\frac{1}{x})}^2=20$$ ($$\because$$ $$x^2+\frac{1}{x^2}=18$$)

$$\Rightarrow x+\frac{1}{x}=\pm \sqrt{20}$$               (Taking square root of both sides)

Now, let us consider the following expression: 

$$x-\frac{1}{x}$$ 

Squaring the above expression, we get: 

$${(x-\frac{1}{x})}^2=x^2-2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=x^2-2+\frac{1}{x^2}[(a-b{)}^2=a^2+b^2-2ab]$$

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

$$\Rightarrow {(x-\frac{1}{x})}^2=16$$ ($$\because$$ $$x^2+\frac{1}{x^2}=18$$)

$$\Rightarrow x-\frac{1}{x}=\pm 4$$                  (Taking square root of both sides)