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

- Oct 18th 2011, 05:20 PMChaimHow do you solve a quadratic function when 'c' has a variable?
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 - Oct 18th 2011, 05:26 PMskeeterRe: How do you solve a quadratic function when 'c' has a variable?
- Oct 18th 2011, 05:30 PMChaimRe: How do you solve a quadratic function when 'c' has a variable?
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. - Oct 18th 2011, 05:36 PMskeeterRe: How do you solve a quadratic function when 'c' has a variable?
if the vertex of a parabola is on the x-axis, then there is a single zero of multiplicity two ... in that case, the discriminant = 0

- Oct 18th 2011, 05:41 PMChaimRe: How do you solve a quadratic function when 'c' has a variable?
- Oct 18th 2011, 05:45 PMskeeterRe: How do you solve a quadratic function when 'c' has a variable?
$\displaystyle y = x^2 + bx + 3b$

$\displaystyle a = 1$ , $\displaystyle b = b$ , $\displaystyle c = 3b$

$\displaystyle b^2 - 4ac = 0$

$\displaystyle b^2 - 4(1)(3b) = 0$

$\displaystyle b^2 - 12 b = 0$

$\displaystyle b(b - 12) = 0$

see it now? - Oct 18th 2011, 05:48 PMChaimRe: How do you solve a quadratic function when 'c' has a variable?
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? - Oct 18th 2011, 05:58 PMskeeterRe: How do you solve a quadratic function when 'c' has a variable?
- Oct 18th 2011, 06:08 PMChaimRe: How do you solve a quadratic function when 'c' has a variable?
- Oct 18th 2011, 06:24 PMChaimRe: How do you solve a quadratic function when 'c' has a variable?
- Oct 19th 2011, 10:30 AMskeeterRe: How do you solve a quadratic function when 'c' has a variable?
- Oct 20th 2011, 02:24 PMChaimRe: How do you solve a quadratic function when 'c' has a variable?
- Oct 22nd 2011, 04:58 PMChaimRe: How do you solve a quadratic function when 'c' has a variable?
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.