Question:easy

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
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Use the general shortcut.
For any non zero $x$, if $x + \dfrac{1}{x} = k$ then $x^{2} + \dfrac{1}{x^{2}} = k^{2} - 2$.
The $-2$ appears because squaring $x + \dfrac{1}{x}$ throws in a cross term of $2 \times x \times \dfrac{1}{x} = 2$, which must be taken back out.

Step 2: Apply it with k equal to 3.
\[ x^{2} + \frac{1}{x^{2}} = 3^{2} - 2 = 9 - 2 = 7 \]

Step 3: A second route using the quadratic.
Multiply the given equation by $x$:
\[ x^{2} + 1 = 3x \]
so $x^{2} = 3x - 1$. Divide the same relation by $x^{2}$ to get $\dfrac{1}{x^{2}} = \dfrac{3}{x} - 1$.
Adding the two:
\[ x^{2} + \frac{1}{x^{2}} = 3x - 1 + \frac{3}{x} - 1 = 3\left(x + \frac{1}{x}\right) - 2 \]
\[ = 3(3) - 2 = 7 \]
Two independent routes give 7, so the result is solid.

Step 4: Check the choices.
9 ignores the cross term and is the commonest slip.
10 would need $k^{2} + 1$, a formula that does not exist.
27 looks like $3^{3}$ and belongs to no correct identity here.

Step 5: One useful remark.
Since $k = 3$ is bigger than 2, real values of $x$ do exist. The value $x^{2} + \dfrac{1}{x^{2}} = 7$ is also greater than 2, as it must be by the same reasoning.

Final Answer:
The value is 7, option (D). \[ \boxed{7} \]
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