Question

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

Answer 1

First one

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

$\sqrt{b^2-4ac}$

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
$\sqrt{b^2-4ac}$<0
we have
1x^2-6x+c=0
a=1
b=-6
c=c
$\sqrt{(-6)^2-4(1)(c)}$<0
$\sqrt{36-4c}$<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