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If $x^{2}+\frac{1}{x^{2}}=18$, find the values of $x+\frac{1}{x}$ and $x-\frac{1}{x}$.
Given:
$x^2 + \frac{1}{x^2} = 18$
To do:
We have to find the values of $x + \frac{1}{x}$ and $x - \frac{1}{x}$.
Solution:
The given expression is $x^2 + \frac{1}{x^2} = 18$. Here, we have to find the values of $x + \frac{1}{x}$ and $x - \frac{1}{x}$. So, by using the identities $(a+b)^2=a^2+2ab+b^2$...................(i) and $(a-b)^2=a^2-2ab+b^2$.............(ii), we can find the required values.
Now,
$x^2 + \frac{1}{x^2} = 18$
Adding $2$ on both sides, we get,
$x^2 + \frac{1}{x^2} + 2 = 18+2$
$x^2 + \frac{1}{x^2} + 2 \times x \times \frac{1}{x} = 20$ (Since $2\times x \times \frac{1}{x}=2$)
$(x+\frac{1}{x})^2=20$ [Using (i)]
Taking square root on both sides, we get,
$x+\frac{1}{x}=\sqrt{20}$
Now,
$x^2 + \frac{1}{x^2} = 18$
Subtracting $2$ on both sides, we get,
$x^2 + \frac{1}{x^2} - 2 = 18-2$
$x^2 + \frac{1}{x^2} - 2 \times x \times \frac{1}{x} = 16$ (Since $2\times x \times \frac{1}{x}=2$)
$(x-\frac{1}{x})^2=16$ [Using (ii)]
Taking square root on both sides, we get,
$x-\frac{1}{x}=\sqrt{16}$
$x-\frac{1}{x}=4$
Hence, the value of $x+\frac{1}{x}$ is $\sqrt{20}$ and the value of $x-\frac{1}{x}$ is $4$.
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