Determine the type and number of solutions of 3x^2 + 6x + 4 = 0. a) one real solution and one imaginary solution b) one real solution c) two real solutions d) two imaginary solutions

D

given a quadratic in standard form : ax² + bx + c = 0

To determine the nature of the roots use the discriminant : b² - 4ac

• If b² - 4ac > 0 , then two real and distinct roots

• If b² - 4ac = 0 , then real and equal root

• I f b² - 4ac < 0 , then roots are not real

here a = 3, b = 6 and c = 4

b² - 4ac = 6² - ( 4 × 3 × 4 ) = 36 - 48 = - 12

Hence 2 imaginary solutions

47 - 90 = 43

43 - 180 = x

x = 137

x-3y=4

step-by-step e