If and are the roots of the equation 3x2 + bx 2 = 0. Find the value of \(\frac{1}{\alpha}\) + \(\frac{1}{\beta}\)
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Correct Answer: Option D
Explanation:
\(\frac{1}{\alpha}\) + \(\frac{1}{\beta}\) = \(\frac{\beta -\alpha}{\alpha \beta}\)
3x2 + 5x + 5x − 2 = 0.
Sum of root = α + β
Product of root = αβ
x2 + \(\frac{5x}{3}\) − \(\frac{2}{3}\) = 0
αβ = − \(\frac{-2}{3}\)
α + β = \(\frac{5}{3}\)
∴ \(\frac{\alpha + \beta}{\alpha \beta}\) = − \(\frac{\frac{5}{3}}{\frac{2}{3}}}\)
= − \(\frac{2}{3}\) × \(\frac{3}{3}\)
= \(\frac{5}{2}\)
\(\frac{1}{\alpha}\) + \(\frac{1}{\beta}\) = \(\frac{\beta -\alpha}{\alpha \beta}\)
3x2 + 5x + 5x − 2 = 0.
Sum of root = α + β
Product of root = αβ
x2 + \(\frac{5x}{3}\) − \(\frac{2}{3}\) = 0
αβ = − \(\frac{-2}{3}\)
α + β = \(\frac{5}{3}\)
∴ \(\frac{\alpha + \beta}{\alpha \beta}\) = − \(\frac{\frac{5}{3}}{\frac{2}{3}}}\)
= − \(\frac{2}{3}\) × \(\frac{3}{3}\)
= \(\frac{5}{2}\)