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Q.

If $(x - \frac{1}{x}) = 4,$ then the value of $(x^{2} + \frac{1}{x^{2}})$ =[[1]].

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answer is 18.

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Detailed Solution

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If

(x - \frac{1}{x}) = 4,

then the value of

(x^{2} + \frac{1}{x^{2}})

= 18.
Here, we are given the value of an algebraic expression as follows,

(x - \frac{1}{x}) = 4

.
Now, it can be observed that the expression

(x^{2} + \frac{1}{x^{2}})

, has both terms squared.
So, to find its value, we will have to square the equation

(x - \frac{1}{x}) = 4

, and then solve to get the required value. So, we have,

\begin{array}{l} {(x - \frac{1}{x})}^{2} = (^{4) 2} \\ {(x - \frac{1}{x})}^{2} = 16 \end{array}

Now, we know the following identity,

{(a - b)}^{2} = a^{2} + b^{2} - 2 a b

,

\begin{array}{l} (^{x) 2} + {(\frac{1}{x})}^{2} - 2 \times x \times \frac{1}{x} = 16 \\ x^{2} + \frac{1}{x^{2}} - 2 = 16 \end{array}

\begin{array}{l} x^{2} - 2 + \frac{1}{x^{2}} + 2 = 16 + 2 \\ x^{2} + \frac{1}{x^{2}} = 18 \end{array}

Hence, the value of

(x^{2} + \frac{1}{x^{2}})

is 18.

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