Question:

If \( x + \dfrac{1}{x} = 3 \), then \( x^{2} + \dfrac{1}{x^{2}} \) will be

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Square both sides. The cross term 2 times x times 1/x is just 2, so x^2 + 1/x^2 = 9 - 2.
Updated On: Jul 17, 2026
  • 9
  • 10
  • 27
  • 7
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The Correct Option is D

Solution and Explanation

Step 1: Pick the right identity.
The square of a sum is \[ (a + b)^{2} = a^{2} + 2ab + b^{2} \]
Take \( a = x \) and \( b = \dfrac{1}{x} \). The middle term becomes \( 2 \cdot x \cdot \dfrac{1}{x} = 2 \), a plain number. That cancellation is exactly why this identity is the natural tool here.

Step 2: Square both sides of the given equation.
\[ \left(x + \frac{1}{x}\right)^{2} = 3^{2} \]
Expanding the left side:
\[ x^{2} + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^{2}} = 9 \]
\[ x^{2} + 2 + \frac{1}{x^{2}} = 9 \]

Step 3: Move the constant across.
\[ x^{2} + \frac{1}{x^{2}} = 9 - 2 = 7 \]

Step 4: Confirm with the actual root.
From \( x + \frac{1}{x} = 3 \), multiply by \( x \) to get \( x^{2} - 3x + 1 = 0 \), so \( x = \dfrac{3 \pm \sqrt{5}}{2} \).
Taking \( x = \dfrac{3 + \sqrt{5}}{2} \approx 2.618 \), we get \( x^{2} \approx 6.854 \) and \( \dfrac{1}{x^{2}} \approx 0.146 \). Their sum is about 7. The identity route is far quicker but the numbers agree.

Step 5: Why the other options are wrong.
9 is simply \( 3^{2} \), which is what you get if you square 3 and forget the middle term \( +2 \). This is the trap option.
10 comes from adding 1 instead of subtracting 2.
27 is \( 3^{3} \), which answers a different question, \( x^{3} + \dfrac{1}{x^{3}} \), and even that is 18, not 27.

Final Answer:
Subtracting the 2 from 9 leaves 7. \[ \boxed{7} \]
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