Question

If $\frac{5(4x^2+1)-3x}{4x}=8$ and x ≠ 0, then what is the value of $(√2x-\frac{1}{√2x})$?

• A. $\sqrt{(\frac{3}{2})}$
• B. $\sqrt{(\frac{5}{2})}$
• C. $\sqrt{(\frac{5}{4})}$
• D. $\sqrt{(\frac{3}{4})}$

Answer 1

Answer: (A)

$\sqrt{(\frac{3}{2})}$

Answer 2

The correct option is (A):$\sqrt{(\frac{3}{2})}$.
$\frac{5(4x^2+1)-3x}{4x}=8$
20x2+ 5 - 3x = 32x
20x2+ 5 = 35x
4x2- 7x + 1 = 0
4x2+ 1 = 7x... (i)
$Now, (\sqrt2x-\frac{1}{\sqrt2x})=2x-2+\frac{1}{2x}$
=$\frac{ (4x2- 4x + 1)}{2x}$
= $\frac{(7x - 4x)}{2x}$ (from (i))
Therefore, $(\sqrt2x-\frac{1}{\sqrt2x})=\sqrt\frac{3}{2}$