# 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 $4$.

Updated on: 01-Apr-2023

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