Determine if the quadratic equation has 2 solutions, one, or 2 imaginary solutions by using the descriminant

Hello from MrBillDoesMath!

31a.  2 real roots

31b   2 complex roots

Discussion:

31a.

2x^2 + 3x - 6 = 0

Using standard quadratic notation, the above equation has a  = 2, b = 3 and c = -6. The discriminant is given by b^2 - 4ac, which is

3^2 - 4*2*(-6) = 9 + 48 = 57 > 0, so the quadratic has 2 real roots

31b.

The equation is equivalent to  16x^2 + 10x  +3 = 0. Its discriminant is

10^2 - 4 * (16) * (3) = 100 - 192 = -92 so this equation has two complex roots

Regards,  

MrB

step-by-step explanation: