If \(x+\frac{1}{x}=3\), then the value of \(x^2+\frac{1}{x^2}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 7
Squaring \(x+\frac{1}{x}\) gives \(x^2+\frac{1}{x^2}\) plus a middle term \(2\times x\times\frac{1}{x}\), which is always 2. So the required value is the square of 3 less 2.
\(\displaystyle \left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2\)
\(\displaystyle 9=x^2+\frac{1}{x^2}+2\)
\(\displaystyle x^2+\frac{1}{x^2}=7\)
The 9 in Option (1) is \(3^2\) with the middle term 2 left in.