Exams › SSC CGL (Prelims) › Maths

If $x+\frac{1}{x}=3$, then the value of $9x^2+\frac{1}{x^2}-1$ is:

1. 87
2. 81
3. 77
4. 79

Correct answer: 81

Solution

From $x+\frac{1}{x}=3$, squaring gives $x^2+\frac{1}{x^2}=7$. Then $9x^2+\frac{1}{x^2}-1=8x^2+\left(x^2+\frac{1}{x^2}\right)-1$. Substituting the identity leads to 81.

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