If \( \alpha, \beta \) are roots of the equation \( 12x^2 - 20x + 3 = 0 \), \( \lambda \in \mathbb{R} \). If \( \frac{1}{2} \leq |\beta - \alpha| \leq \frac{3}{2} \), then the sum of all possible values of \( \lambda \) is:
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Correct Answer: 3
Solution and Explanation
The given quadratic equation is:
\(12x^2 - 20x + 3 = 0\)
Step 2: Find the roots using the quadratic formulaFor a quadratic equation \(ax^2 + bx + c = 0\), the roots are given by:
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Here, \(a = 12\), \(b = -20\), and \(c = 3\).
Discriminant, \(D = b^2 - 4ac = (-20)^2 - 4(12)(3) = 400 - 144 = 256\)
\(\sqrt{D} = 16\)
\(\alpha = \dfrac{20 + 16}{24} = \dfrac{36}{24} = \dfrac{3}{2}\)
\(\beta = \dfrac{20 - 16}{24} = \dfrac{4}{24} = \dfrac{1}{6}\)
Step 3: Find the absolute difference of the roots\(|\beta - \alpha| = \left|\dfrac{1}{6} - \dfrac{3}{2}\right|
= \left|\dfrac{1 - 9}{6}\right| = \dfrac{8}{6} = \dfrac{4}{3}
Step 4: Apply the given condition\(\dfrac{1}{2} \leq |\beta - \alpha| \leq \dfrac{3}{2}\)
Since \(\dfrac{1}{2} \leq \dfrac{4}{3} \leq \dfrac{3}{2}\), the condition is satisfied.
Final Answer:\(\boxed{7.00}\)
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