Question
Evaluate the following using identities:$\Big(2\text{x}-\frac{1}{\text{x}}\Big)^2$
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Answer
We have,$\Big(2\text{x}-\frac{1}{\text{x}}\Big)^2=(2\text{x})^2+\big(\frac{1}{\text{x}}\big)^2-2.2\text{x}.\frac{1}{\text{x}}$
$\Big(2\text{x}-\frac{1}{\text{x}}\Big)^2=4\text{x}^2+\frac{1}{\text{x}^2}-4$ $\begin{bmatrix}\because(\text{a}-\text{b})^2=\text{a}^2+\text{b}^2-2\text{ab},\\\text{Where}\ \text{a}=2\text{x}\ \text{and}\ \text{b}=\frac{1}{\text{x}}\end{bmatrix}$
$\therefore\Big(2\text{x}-\frac{1}{\text{x}}\Big)^2=4\text{x}^2+\frac{1}{\text{x}^2}-4$
$\Big(2\text{x}-\frac{1}{\text{x}}\Big)^2=4\text{x}^2+\frac{1}{\text{x}^2}-4$ $\begin{bmatrix}\because(\text{a}-\text{b})^2=\text{a}^2+\text{b}^2-2\text{ab},\\\text{Where}\ \text{a}=2\text{x}\ \text{and}\ \text{b}=\frac{1}{\text{x}}\end{bmatrix}$
$\therefore\Big(2\text{x}-\frac{1}{\text{x}}\Big)^2=4\text{x}^2+\frac{1}{\text{x}^2}-4$
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