determine if the quadratic equation has 2 solutions, one, or 2 imaginary solutions by using the descriminant
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Hello from MrBillDoesMath!
Answer:
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