How do you solve a quadratic function when 'c' has a variable?

**Chaim** · October 18th 2011, 05:20 PM

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

**skeeter** · October 18th 2011, 05:26 PM

Originally Posted by Chaim

Well the equation is a^2 + ba + 3a
So I'm having trouble with this, cause the last thing has a variable
I'm suppose to find the vertex and what the possible values of a could be.

what you have posted is an expression, not an equation. you cannot "solve" an expression. note that an equation has an equal sign ...

$a^2 + ba + 3a = something$

also ... are you missing the variable "x" in the first two terms?

**Chaim** · October 18th 2011, 05:30 PM

Originally Posted by skeeter

what you have posted is an expression, not an equation. you cannot "solve" an expression. note that an equation has an equal sign ...

$a^2 + ba + 3a = something$

also ... are you missing the variable "x" in the first two terms?

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.

**skeeter** · October 18th 2011, 05:36 PM

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

**Chaim** · October 18th 2011, 05:41 PM

Originally Posted by skeeter

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

Thank you, I also got the answer book, but I don't know how to work it out, the answer that 'b' could be is 0 or 12
How could 'b' be 12 too?

**skeeter** · October 18th 2011, 05:45 PM

$y = x^2 + bx + 3b$

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

$b^2 - 4ac = 0$

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

$b^2 - 12 b = 0$

$b(b - 12) = 0$

see it now?

**Chaim** · October 18th 2011, 05:48 PM

Originally Posted by skeeter

$y = x^2 + bx + 3b$

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

$b^2 - 4ac = 0$

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

$b^2 - 12 b = 0$

$b(b - 12) = 0$

see it now?

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?

**skeeter** · October 18th 2011, 05:58 PM

Originally Posted by Chaim

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?

$b^2-4ac$ ...

is the discriminant, and yes, it's the expression under the radical in the quadratic formula ... I recommend you research it and find out what its value can tell you about zeros of a quadratic function.

**Chaim** · October 18th 2011, 06:08 PM

Originally Posted by skeeter

$b^2-4ac$ ...

is the discriminant, and yes, it's the expression under the radical in the quadratic formula ... I recommend you research it and find out what its value can tell you about zeros of a quadratic function.

Thank you very much

**Chaim** · October 18th 2011, 06:24 PM

Originally Posted by skeeter

$b^2-4ac$ ...

is the discriminant, and yes, it's the expression under the radical in the quadratic formula ... I recommend you research it and find out what its value can tell you about zeros of a quadratic function.

Yeah.. think you could help me again?
It's basically the same problem, but it says "What if x^2 + 3bx - b^2 + 1"
I am confused with that because the radical formula wouldn't work here then.
Same vertex.

**skeeter** · October 19th 2011, 10:30 AM

Originally Posted by Chaim

Yeah.. think you could help me again?
It's basically the same problem, but it says "What if x^2 + 3bx - b^2 + 1"
I am confused with that because the radical formula wouldn't work here then.
Same vertex.

$a = 1$ , $b = 3b$ , $c = -b^2 + 1$

substitute these values into the discriminant equation ...

$b^2 - 4ac = 0$

... solve for the value b

**Chaim** · October 20th 2011, 02:24 PM

Originally Posted by skeeter

$a = 1$ , $b = 3b$ , $c = -b^2 + 1$

substitute these values into the discriminant equation ...

$b^2 - 4ac = 0$

... solve for the value b

Oh ok! Thanks I get it now!
At first I was confuse with 'c' since there was a variable, and a number remaining
But I see that you just put them together as 'c'!
Ok thanks

**Chaim** · October 22nd 2011, 04:58 PM

Originally Posted by skeeter

$a = 1$ , $b = 3b$ , $c = -b^2 + 1$

substitute these values into the discriminant equation ...

$b^2 - 4ac = 0$

... solve for the value b

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.

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