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

Do the math. she originally had 50 dollars on her gift card and she had 12 dollars left on her gift card after she finished her shopping spree. if you subtract 50 minus 12 you get thirty eight dollars as your answer. so therefore, ciara spent 38 dollars total o