Question:easy

If \[ x+\frac1x=3, \] then the value of \[ x^2+\frac1{x^2} \] is:

Show Hint

Memorize the identity \[ \left(x+\frac1x\right)^2 = x^2+\frac1{x^2}+2. \] It is one of the most frequently used identities in algebra.
Updated On: Jun 10, 2026
  • \(5\)
  • \(7\)
  • \(9\)
  • \(11\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Write the given relation.
We know $x+\dfrac{1}{x}=3$. We must find $x^2+\dfrac{1}{x^2}$.

Step 2: Pick the right identity.
Squaring a sum gives $\left(x+\dfrac{1}{x}\right)^2=x^2+\dfrac{1}{x^2}+2$, because the middle term $2\cdot x\cdot\dfrac{1}{x}=2$.

Step 3: Square the given value.
Since $x+\dfrac{1}{x}=3$, squaring both sides gives \[ \left(x+\frac{1}{x}\right)^2=3^2=9. \]

Step 4: Expand the left side.
Using the identity, \[ x^2+\frac{1}{x^2}+2=9. \]

Step 5: Move the $2$ across.
Subtract $2$ from both sides. \[ x^2+\frac{1}{x^2}=9-2=7. \]

Step 6: State the answer.
So the required value is $7$. We never needed to actually solve for $x$. \[ \boxed{7} \]
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