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

1 $\frac{19}{4}$ $\frac{{19}}{4}$
2 $\frac{23}{4}$ $\frac{{23}}{4}$
3 $\frac{27}{4}$ $\frac{{27}}{4}$
4 $\frac{31}{4}$ $\frac{{31}}{4}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} 2 x - \frac{1}{2 x} = 5 \\ Divide both side \\ x - \frac{1}{4 x} = \frac{5}{2} \\ Squaring both side \\ \Rightarrow x^{2} + \frac{1}{16 x^{2}} - 2 \times x \times \frac{1}{4 x} = \frac{25}{4} \\ \Rightarrow x^{2} + \frac{1}{16 x^{2}} - \frac{1}{2} = \frac{25}{4} \\ \Rightarrow x^{2} + \frac{1}{16 x^{2}} = \frac{25}{4} + \frac{1}{2} \\ \Rightarrow x^{2} + \frac{1}{16 x^{2}} = \frac{27}{4} \\ So, \\ x^{2} + \frac{1}{16 x^{2}} - 2 \\ = \frac{27}{4} - 2 \\ = \frac{19}{4} \end{aligned}

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

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\begin{aligned} 2 x - \frac{1}{2 x} = 5 \\ Divide both side \\ x - \frac{1}{4 x} = \frac{5}{2} \\ Squaring both side \\ \Rightarrow x^{2} + \frac{1}{16 x^{2}} - 2 \times x \times \frac{1}{4 x} = \frac{25}{4} \\ \Rightarrow x^{2} + \frac{1}{16 x^{2}} - \frac{1}{2} = \frac{25}{4} \\ \Rightarrow x^{2} + \frac{1}{16 x^{2}} = \frac{25}{4} + \frac{1}{2} \\ \Rightarrow x^{2} + \frac{1}{16 x^{2}} = \frac{27}{4} \\ So, \\ x^{2} + \frac{1}{16 x^{2}} - 2 \\ = \frac{27}{4} - 2 \\ = \frac{19}{4} \end{aligned}

\begin{aligned} 2 x - \frac{1}{2 x} = 5 \\ Divide both side \\ x - \frac{1}{4 x} = \frac{5}{2} \\ Squaring both side \\ \Rightarrow x^{2} + \frac{1}{16 x^{2}} - 2 \times x \times \frac{1}{4 x} = \frac{25}{4} \\ \Rightarrow x^{2} + \frac{1}{16 x^{2}} - \frac{1}{2} = \frac{25}{4} \\ \Rightarrow x^{2} + \frac{1}{16 x^{2}} = \frac{25}{4} + \frac{1}{2} \\ \Rightarrow x^{2} + \frac{1}{16 x^{2}} = \frac{27}{4} \\ So, \\ x^{2} + \frac{1}{16 x^{2}} - 2 \\ = \frac{27}{4} - 2 \\ = \frac{19}{4} \end{aligned}

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

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If 2x−12x=5,2x−12x=5,{\text{2}}x - \frac{1}{{2x}} = 5{\text{,}} x≠0,x≠0,{\text{x}} \ne {\text{0,}} then find the value of x2+116x2−2x2+116x2−2{x^2} + \frac{1}{{16{x^2}}} - 2 = ?

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

NCERT
6 to 12