To solve the problem, we need to find the value of \(\left(a - \frac{1}{a}\right)\) given that \(\left(a^2 + \frac{1}{a^2}\right) = 18\).
We begin by using the identity:
\(a^2 + \frac{1}{a^2} = \left(a - \frac{1}{a}\right)^2 + 2.\)
Given that \(a^2 + \frac{1}{a^2} = 18\), we can write:
\(\left(a - \frac{1}{a}\right)^2 + 2 = 18\)
Simplifying the equation, we get:
\(\left(a - \frac{1}{a}\right)^2 = 18 - 2\)
\(\left(a - \frac{1}{a}\right)^2 = 16\)
Taking the square root on both sides, we find:
\(\left(a - \frac{1}{a}\right) = \pm 4\)
The problem does not specify whether \(a\) is positive or negative, but typically in such questions, we consider the principal (positive) value. Therefore,:
\(a - \frac{1}{a} = 4\)
Thus, the correct answer is 4.
Based on the analysis and calculation above, the correct option is 4.
Similar Questions
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If $a + b + c = 0$, then find the value of $\frac{(a^2+b^2+c^2)^2}{a^2b^2+b^2c^2+c^2a^2}$
- If $x^4 + \frac{1}{x^4} = 322$, then $x^3 - \frac{1}{x^3} = ?$
- If $(x + \frac{1}{x}) = 7$, then $(x - \frac{1}{x})$ is equal to:
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$(1 + x)(1 + x^2)(1 + x^4)(1 + x^8)(1 - x)$ - If $a + b + c = 17$, $abc = 168$, and $ab + bc + ca = 94$, then $a^3 + b^3 + c^3 = ?$
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Important Questions from Identities
If \(2x + { {1} \over 3x}= 5, x ≠ 0\) , then what is the value of \(27x^3+{{1} \over 8x^3}\) ?
If x 2 + 4y 2 = 40, xy = 6 and x > 2y then the value of x - 2y is:
- \(\dfrac{(0.73)^3+(0.31)^3}{(0.73)^2-0.73\times0.31+(0.31)^2}\)
- \(\dfrac{(5.17-2.19)^2-(5.17+2.19)^2}{11.3223}\)
If a(a + b + c) 2= 1792; b(a + b + c) 2= 1536; c(a + b + c) 2= 768 then what will be the value of a?
