Well the equation is x^2 + bx + 3b
So I'm having trouble with this, cause the last thing has a variable
I'm suppose to find the value of b
Vertex is on the x-axis
What if x^2 + 3bx - b^2 + 1
Well the equation is x^2 + bx + 3b
So I'm having trouble with this, cause the last thing has a variable
I'm suppose to find the value of b
Vertex is on the x-axis
What if x^2 + 3bx - b^2 + 1
Oh, just imagine that the 'a' are the 'x'
What I forgot to mention was that it has its vertex on the x-axis, so I'm suppose to find the value of b
Ahh... I think I wrote it wrong
So I'll rewrite it with 'x' instead of 'a' so the equation is x^2 + bx + 3b
So I have to find the value of b.
Oh I see thanks!
By the way, what equation was b^2 - 4ac = 0?
Did that came from the quadratic formula which is -b + or - square root of b^2 - 4ac all over 2?
And to get 0, you did f(x) = 0
so f(0) = (0)^2 + b(0) + 3d
0 = 3d
Divide both side by 3
0 = d
Did I do that right?
Just wondering, could you help me again, but in a different way?
My teacher really didn't show us using the 'discriminant way' but in a different way
Just making sure, in case if I can't use the discriminant
This is what I did to start off with f(x) = x^2 + bx + 3b
When vertex is on the x-axis, y=0
(-b/2a) = (-b/2) = x
0=(-b/2)^2 + b(-b/2)+3b
0=(b^2/4) + -b^2/2 + 3b
Then I multipled everything by 4
0=b^2 - (4b^2/2) + 12b = 0
Then I got lost after this part, cause the 'b' part is weird so it's tough to do the quadratic equation on this.