Question:

If the roots of the equation \( 3x^2 + 4kx + 3 = 0 \) are non-real, then \( k \) lies in the interval:

Show Hint

To determine when a quadratic has non-real roots, always apply the discriminant condition \( b^2 - 4ac<0 \). Then solve the resulting inequality.
Updated On: May 13, 2025
  • \( \left[-2, -\frac{3}{2} \right] \)
  • \( \left[\frac{3}{2}, 2 \right] \)
  • \( \left( -\frac{3}{2}, \frac{3}{2} \right) \)
  • \( (2, 3) \)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

We know that the roots are non-real \( \Rightarrow \) discriminant \( D<0 \). \[ D = (4k)^2 - 4ac = 16k^2 - 36<0 \Rightarrow 16k^2<36 \Rightarrow k^2<\frac{9}{4} \] \[ \Rightarrow -\frac{3}{2}<k<\frac{3}{2} \]
Was this answer helpful?
0
0