Question

If ax2 + bx + 6 = 0 doesn't have 2 distinct real roots, then find the least value of 3a + b​

Answer 1

Find an answer to your question if ax^2 + bx + 6=0 doesn't have 2 distinct real roots, then find the least value of 3a +

Answer 2

Answer:

Since The QE has no distinct roots, so b^2 = 24a

Or, 3a = b^2 / 8

So 3a+b = b + b^2/8 =(8b+b^2)/8 = {(b+4)^2 - 16}/8 = (b+4)^2/8 - 2.

The minimum value of this will clearly be -2 -------------(when b = -4 and 3a = 2, or a=2/3)

Hope this will help you ☺️♥️