3(x+1)= -2(x-1)-4 I dont necessarily want the answer, I just want the steps.
2 answers:
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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
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
we have
1x^2-6x+c=0
a=1
b=-6
c=c
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