In the equation x^2-6x+c=0, find the values of c that will give two imaginary solutions.

a. c > = 5
b. c > 6
c. c < 6
d. c > 9

i figured out the equation with 6 (so i believe either b or c is the answer) being there, but is would it be > or < ?

rewrite x^2+9 as a product using complex numbers and the polynomial identity below.

a. (x+3)(x-3)
b. (x+3i)(x-3i)
c. (3x)(3i)
d. (x+9i)(x-9i)

i'm not sure if it's b or d, if i'm on the right path. ?

First one

when doing the quadratic equation, we come across something called the determinant



when this is positive, we have 2 real roots
when this is 0, we have 1 real root
whenthis is negative, we have 2 imaginary rots
we want 2imaginary roots so
<0
we have
1x^2-6x+c=0
a=1
b=-6
c=c
<0
<0
square root both sies
36-4c<0
add 4c both sides
36<4c
divide both sides by 4
9<c
c>9

D is the answer (not b or c)

remember
√-1=i
a²-b²=(a-b)(a+b)

we want to do
x^2-(-9)
now we have to take the sqrt of -9
√-9=(√-1)(√9)=(i)(3)=3i

(x)^3-(3i)^2
(x-3i)(x+3i)
B is answer