Question:

If \(x+\dfrac{1}{x}=3\) find \(x^{2}+\dfrac{1}{x^{2}}\).

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Square $x+\frac{1}{x}$ to get $x^2+\frac{1}{x^2}$ quickly: $(x+\tfrac{1}{x})^2=x^2+2+\tfrac{1}{x^2}$.
Updated On: Aug 11, 2025
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The Correct Option is D

Solution and Explanation

Given the equation \(x+\dfrac{1}{x}=3\), we need to find \(x^{2}+\dfrac{1}{x^{2}}\).

We use the identity for squares: \(\left(x+\dfrac{1}{x}\right)^2=x^2+2+\dfrac{1}{x^2}\).

First, calculate \(\left(x+\dfrac{1}{x}\right)^2\):

\[\left(3\right)^2=9\]

Therefore,

\[x^2+2+\dfrac{1}{x^2}=9\]

By rearranging, we get:

\[x^2+\dfrac{1}{x^2}=9-2\]

\[x^2+\dfrac{1}{x^2}=7\]

Hence, the solution is \(7\).

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