If $${\text{2}}x - \frac{1}{{2x}} = 5{\text{,}}$$ $${\text{x}} \ne {\text{0,}}$$ then find the value of $${x^2} + \frac{1}{{16{x^2}}} - 2$$ = ? Options: A . &nbsp$$\frac{{19}}{4}$$ B . &nbsp$$\frac{{23}}{4}$$ C . &nbsp$$\frac{{27}}{4}$$ D . &nbsp$$\frac{{31}}{4}$$

If $${\text{2}}x - \frac{1}{{2x}} = 5{\text{,}}$$ $${\text{x}} \ne {\text{0

Question

If $${\text{2}}x - \frac{1}{{2x}} = 5{\text{,}}$$ $${\text{x}} \ne {\text{0,}}$$ then find the value of $${x^2} + \frac{1}{{16{x^2}}} - 2$$ = ?

Options:

A . $$\frac{{19}}{4}$$

B . $$\frac{{23}}{4}$$

C . $$\frac{{27}}{4}$$

D . $$\frac{{31}}{4}$$

Answer: Option A
$$\eqalign{
& 2x - \frac{1}{{2x}} = 5 \cr
& {\text{Divide by 2 both side}} \cr
& x - \frac{1}{{4x}} = \frac{5}{2} \cr
& {\text{Squaring both side}} \cr
& \Rightarrow {x^2} + \frac{1}{{16{x^2}}} - 2 \times x \times \frac{1}{{4x}} = \frac{{25}}{4} \cr
& \Rightarrow {x^2} + \frac{1}{{16{x^2}}} - \frac{1}{2} = \frac{{25}}{4} \cr
& \Rightarrow {x^2} + \frac{1}{{16{x^2}}} = \frac{{25}}{4} + \frac{1}{2} \cr
& \Rightarrow {x^2} + \frac{1}{{16{x^2}}} = \frac{{27}}{4} \cr
& {\text{So, }} \cr
& {x^2} + \frac{1}{{16{x^2}}} - 2 \cr
& = \frac{{27}}{4} - 2 \cr
& = \frac{{19}}{4} \cr} $$

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