Answer:
d.
[tex]3 {x}^{2} + 4x + 2 = 0[/tex]
has no real roots.
Step-by-step explanation:
A quadratic equation has no real solution if:
[tex]{ {b}^{2} - 4ac } < 0[/tex]
a) a = 4 , c = -64, b = 0
therefore, 0 - 4(4 × -64)
= -4(-256)
= 1024 which is greater than 0.
Hence, 4x^ - 64 = 0 has real roots
b) a = 1, b = -4 and c = -8
b^2 - 4ac = (-4)^2 - 4(1 × -8)
= 16 -4(-8)
= 16 + 32
= 48 which is greater than 0.
Hence, x^2 - 4x - 8 = 0 has real roots.
c) a = 2, b = -4 and c = 2
b^2 - 4ac = (-4)^2 - 4(2 × 2)
= 16 - 4(4)
= 16 - 16
= 0
Hence, 2x^2 - 4x + 2 has repeated roots.
d) a = 3, b = 4 and c = 2
b^2 - 4ac = 4^2 - 4(3 × 2)
= 16 - 4(6) = 16 - 24
= -8 which is less than 0.
Therefore,
[tex]3 {x}^{2} + 4x + 2 = 0[/tex]
has no real roots.