Question:
If \(x + \dfrac{1}{x} = 4\), then what is the value of \(x^2 + \dfrac{1}{x^2}\)?
If \(x + \dfrac{1}{x} = 4\), then what is the value of \(x^2 + \dfrac{1}{x^2}\)?
Show Hint
Square the given expression; the cross term is 2.
Updated On: Jul 8, 2026
- 18
- 12
- 16
- 14
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Square both sides.
\[\left(x + \dfrac{1}{x}\right)^2 = 4^2 = 16\] Step 2: Expand the left side.
\[x^2 + 2 + \dfrac{1}{x^2} = 16\] Step 3: Isolate the required term.
\[x^2 + \dfrac{1}{x^2} = 16 - 2 = \boxed{14}\]
\[\left(x + \dfrac{1}{x}\right)^2 = 4^2 = 16\] Step 2: Expand the left side.
\[x^2 + 2 + \dfrac{1}{x^2} = 16\] Step 3: Isolate the required term.
\[x^2 + \dfrac{1}{x^2} = 16 - 2 = \boxed{14}\]
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