Question

If 3x^2+4kx +1 >0 for all real values of x , then k lies in the interval

Answer 1

3x² + 4kx + 1 > 0 for all real value of x
this is possible only when,
D = b² -4ac< 0

( 4 k)² - 4(3) < 0

16K² -12 < 0

4K² -3 < 0

(2K-√3)(2K+√3) < 0

-√3/2 < K < √3/2

Answer 2

If the discriminant is negative then the roots of the quadratic function are imaginary.

So the quadratic function is never 0 for any real x.

Then as the coefficient of x^2 is positive , the value of the quadratic is positive for some large x for sure. So it will be positive for all x.

Discriminant = 16 k^2 - 12 < 0
So. The range of k is
- sqrt(3)/2 < k < sqrt(3)/2
Answer.