Question

If the quadratic equation \( 3x^2 + (2k + 1)x - 5k = 0 \) has real and equal roots, then the value of \( k \) such that \( -\frac{1}{2}<k<0 \) is:

• A. \(\frac{-16 + \sqrt{255}}{2}\)
• B. \(\frac{-16 - \sqrt{255}}{2}\)
• C. \(\frac{-2}{3}\)
• D. \(\frac{-3}{5}\)

Hint:

For real and equal roots, always check if the discriminant is zero and solve for the required parameter.

Answer 1

The Correct Option is A

Step 1: Condition for real and equal roots.
For a quadratic equation \( ax^2 + bx + c = 0 \), real and equal roots occur when the discriminant is zero: \[ \Delta = b^2 - 4ac = 0 \] Substituting \( a = 3 \), \( b = 2k + 1 \), and \( c = -5k \): \[ (2k + 1)^2 - 4(3)(-5k) = 0 \] Step 2: Solving for \( k \).
Expanding and simplifying: \[ 4k^2 + 4k + 1 + 60k = 0 \] \[ 4k^2 + 64k + 1 = 0 \] Solving for \( k \) using the quadratic formula: \[ k = \frac{-64 \pm \sqrt{4096 - 4}}{8} = \frac{-64 \pm \sqrt{255}}{8} \] \[ k = \frac{-16 \pm \sqrt{255}}{2} \] Since \( -\frac{1}{2}<k<0 \), we choose \( \frac{-16 + \sqrt{255}}{2} \).