Question

Determine which functions have two real number zeros by calculating the discriminant, b2 – 4ac. Check all that apply. Please 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

Answer 1

By definition, we have that the discribinamte is given by:
 $b ^ 2 - 4ac$
 If the discriminant is greater than zero, then the function has two real roots.
 We then have to calculate the discriminant for each function:

 For$f (x) = x ^ 2 + 6x + 8:$
 $b ^ 2 - 4ac = 6 ^ 2 - 4 (1) (8) b ^ 2 - 4ac = 36 - 32 b ^ 2 - 4ac = 4$
 The function has two real roots.

 For $g (x) = x ^ 2 + 4x + 8$
 $b ^ 2 - 4ac = 4 ^ 2 - 4 (1) (8) b ^ 2 - 4ac = 16 - 32 b ^ 2 - 4ac = -16$
 The function does not have two real roots.

 For $h (x) = x ^ 2 - 12x + 32$
 $b ^ 2 - 4ac = (-12) ^ 2 - 4 (1) (32) b ^ 2 - 4ac = 144 - 128 b ^ 2 - 4ac = 16$
 The function has two real roots.

 For $k (x) = x ^ 2 + 4x - 1$:
 $b ^ 2 - 4ac = (4) ^ 2 - 4 (1) (- 1) b ^ 2 - 4ac = 16 + 4 b ^ 2 - 4ac = 20$
 The function has two real roots.

 For $p (x) = 5x2 + 5x + 4$:
 $b ^ 2 - 4ac = (5) ^ 2 - 4 (5) (4) b ^ 2 - 4ac = 25 - 80 b ^ 2 - 4ac = -55$
 The function does not have two real roots.

 For $t (x) = x ^ 2 - 2x - 15$:
 $b ^ 2 - 4ac = (-2) ^ 2 - 4 (1) (- 15) b ^ 2 - 4ac = 4 + 60 b ^ 2 - 4ac = 64$
 The function has two real roots.

 Answer:
 functions have two real number zeros
 $f (x) = x ^ 2 + 6x + 8 h (x) = x ^ 2 - 12x + 32 k (x) = x ^ 2 + 4x - 1 t (x) = x ^ 2 - 2x - 15$

Answer 2

F(x) = x^2 + 6x + 8
= b^2 - 4ac
= (6)^2 - 4(1)(8)
= 36 - 4(8)
= 36 - 32
= 4

g(x) = x2 + 4x + 8
= b^2 - 4ac
= (4)^2 - 4(1)(8)
= 16 - 4(8)
= 16 - 32
= -16

h(x) = x2 – 12x + 32
= b^2 - 4ac
= (-12)^2 - 4(1)(32)
= 144 - 4(32)
= 144 - 128
= 16

k(x) = x2 + 4x – 1

= (4)^2 - 4(1)(-1)
= 16 - 4(-1)
= 16 + 4
= 20

p(x) = 5x2 + 5x + 4
= b^2 - 4ac
= (5)^2 - 4(5)(4)
= 25 - 4(20)
= 25 - 80
= -55

t(x) = x2 – 2x – 15
= b^2 - 4ac
= (-2)^2 - 4(1)(-15)
= 4 - 4(-15)
= 4 + 60
= 64