Question

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:

Hint:

For quadratic equations, use Vieta's formulas to find the sum and product of roots, then use the given conditions to solve for other parameters.

Answer 1

Correct Answer: 3

Step 1: Write the given quadratic equation
The given quadratic equation is:

\(12x^2 - 20x + 3 = 0\)

Step 2: Find the roots using the quadratic formula
For 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:

\(\boxed{7.00}\)