Determine which functions have two real number zeros by calculating the discriminant, b2 – 4ac. check all that apply. answer f(x) = x2 + 6x + 8 g(x) = x2 + 4x + 8 h(x) = x2 – 12x + 32 k(x) = x2 + 4x – 1 p(x) = 5x2 + 5x + 4 t(x) = x2 – 2x – 15

The discriminant of a quadratic equation is the b^2-4ac portion that the square root is taken of. If the discriminant is negative, then the function has 2 imaginary roots, if the discriminant is equal to 0, then the function has only 1 real root, and finally, if the discriminant is greater than 0, the function has 2 real roots. So let's look at the equations and see which have a positive discriminant. f(x) = x^2 + 6x + 8 6^2 - 4*1*8 36 - 32 = 4 Positive, so f(x) has 2 real roots. g(x) = x^2 + 4x + 8 4^2 - 4*1*8 16 - 32 = -16 Negative, so g(x) does not have any real roots h(x) = x^2 – 12x + 32 -12^2 - 4*1*32 144 - 128 = 16 Positive, so h(x) has 2 real roots. k(x) = x^2 + 4x – 1 4^2 - 4*1*(-1) 16 - (-4) = 20 Positive, so k(x) has 2 real roots. p(x) = 5x^2 + 5x + 4 5^2 - 4*5*4 25 - 80 = -55 Negative, so p(x) does not have any real roots t(x) = x^2 – 2x – 15 -2^2 - 4*1*(-15) 4 - (-60) = 64 Positive, so t(x) has 2 real roots.