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


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 \( 398 \).

       

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