Question:
If \(\frac{5(4x^2+1)-3x}{4x}=8\) and x ≠ 0, then what is the value of \((√2x-\frac{1}{√2x}) \)?
If \(\frac{5(4x^2+1)-3x}{4x}=8\) and x ≠ 0, then what is the value of \((√2x-\frac{1}{√2x}) \)?
Updated On: Sep 13, 2024
- \(\sqrt{(\frac{3}{2})}\)
- \(\sqrt{(\frac{5}{2})}\)
- \(\sqrt{(\frac{5}{4})}\)
- \(\sqrt{(\frac{3}{4})}\)
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The Correct Option is A
Solution and Explanation
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, \((\sqrt2x-\frac{1}{\sqrt2x})=\sqrt\frac{3}{2}\)
\(\frac{5(4x^2+1)-3x}{4x}=8\)
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, \((\sqrt2x-\frac{1}{\sqrt2x})=\sqrt\frac{3}{2}\)
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