Question:
Is\( (\sqrt{x}-\frac{1}{\sqrt{x}})=\sqrt{-5}\) and x≠ 0, then what is the value of \(\frac{3(x^2+3)-6-3x}{4x}\) ?
Is\( (\sqrt{x}-\frac{1}{\sqrt{x}})=\sqrt{-5}\) and x≠ 0, then what is the value of \(\frac{3(x^2+3)-6-3x}{4x}\) ?
Updated On: Sep 13, 2024
- -3
- -2
- 3
- 2
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The Correct Option is A
Solution and Explanation
The correct option is (A): -3
\((\sqrt{x}=\frac{1}{\sqrt{x}})^2=\sqrt-5\)
\((\sqrt{x}=\frac{1}{\sqrt{x}})^2=-5\)
\(x-2+\frac{1}{x}=-5\)
x2 - 2x + 1 = -5x
x2 + 3x + 1 = 0
\(=\frac{3(3x+6)-6-3x}{4x}\)
\(\frac{-3+6-6-3x}{4x}\)
\(\frac{-12x}{4x}\)
\(=-3\)
\((\sqrt{x}=\frac{1}{\sqrt{x}})^2=\sqrt-5\)
\((\sqrt{x}=\frac{1}{\sqrt{x}})^2=-5\)
\(x-2+\frac{1}{x}=-5\)
x2 - 2x + 1 = -5x
x2 + 3x + 1 = 0
\(=\frac{3(3x+6)-6-3x}{4x}\)
\(\frac{-3+6-6-3x}{4x}\)
\(\frac{-12x}{4x}\)
\(=-3\)
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