Question

If $2m-\frac{1}{2m}=2,$ then the value of $(m^2-\frac{1}{16m^2})$ is:

• A. $1$
• B. $\sqrt{2}$
• C. $\sqrt{3}$
• D. $\sqrt{5}$

Answer 1

The correct option is B $\sqrt{2}$

$2m-\frac{1}{2m}=2$

dividing by $2$

$m-\frac{1}{4m}=1$

square both sides,

${(m-\frac{1}{4m})}^2$

$\Rightarrow m^2+\frac{1}{16m}-\frac{1}{2}=1$

$\Rightarrow m^2+\frac{1}{16m^2}=1+\frac{1}{2}$=$\frac{3}{2}$

$=\frac{3}{2}$

Then ${(m+\frac{1}{4m})}^2=m^2+\frac{1}{16m^2}+2\times m\times \frac{1}{4}\Rightarrow {(m+\frac{1}{4m})}^2=\frac{3}{2}+\frac{1}{2}=2$

Then $m+\frac{1}{4m}=\sqrt{2}$

$m^2-\frac{1}{16m^2}=(m+\frac{1}{4m})(m-\frac{1}{4m})=\sqrt{2}\times 1=\sqrt{2}$