# If $$x + \frac{1}{x} =20$$, find the value of $$x^2 + \frac{1}{x^2}$$.

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Given:

$$x + \frac{1}{x} =20$$

To do:

We have to find the value of $$x^2 + \frac{1}{x^2}.$$

Solution:

The given expression is $$x + \frac{1}{x} =20$$. Here, we have to find the value of $$x^2 + \frac{1}{x^2}$$. So, by squaring the given expression and using the identity $$(a+b)^2=a^2+2ab+b^2$$, we can find the value of $$x^2 + \frac{1}{x^2}.$$

$$(a+b)^2=a^2+2ab+b^2$$...................(i)

Now,

$$x + \frac{1}{x} =20$$

Squaring on both sides, we get,

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

$$x^2+2\times x \times \frac{1}{x}+(\frac{1}{x})^2=400$$           [Using (i)]

$$x^2+2+\frac{1}{x^2}=400$$

$$x^2+\frac{1}{x^2}=400-2$$               (Transposing 2 to RHS)

$$x^2+\frac{1}{x^2}=398$$

Hence, the value of $$x^2+\frac{1}{x^2}$$ is $$398$$.