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

• Oct 18th 2011, 05:20 PM
Chaim
How 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 PM
skeeter
Re: How do you solve a quadratic function when 'c' has a variable?
Quote:

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 ...

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

also ... are you missing the variable "x" in the first two terms?
• Oct 18th 2011, 05:30 PM
Chaim
Re: How do you solve a quadratic function when 'c' has a variable?
Quote:

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 ...

\$\displaystyle 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.
• Oct 18th 2011, 05:36 PM
skeeter
Re: 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 PM
Chaim
Re: How do you solve a quadratic function when 'c' has a variable?
Quote:

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?
• Oct 18th 2011, 05:45 PM
skeeter
Re: 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 PM
Chaim
Re: How do you solve a quadratic function when 'c' has a variable?
Quote:

Originally Posted by skeeter
\$\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?

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 PM
skeeter
Re: How do you solve a quadratic function when 'c' has a variable?
Quote:

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?

\$\displaystyle 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.
• Oct 18th 2011, 06:08 PM
Chaim
Re: How do you solve a quadratic function when 'c' has a variable?
Quote:

Originally Posted by skeeter
\$\displaystyle 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
• Oct 18th 2011, 06:24 PM
Chaim
Re: How do you solve a quadratic function when 'c' has a variable?
Quote:

Originally Posted by skeeter
\$\displaystyle 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.
• Oct 19th 2011, 10:30 AM
skeeter
Re: How do you solve a quadratic function when 'c' has a variable?
Quote:

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.

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

substitute these values into the discriminant equation ...

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

... solve for the value b
• Oct 20th 2011, 02:24 PM
Chaim
Re: How do you solve a quadratic function when 'c' has a variable?
Quote:

Originally Posted by skeeter
\$\displaystyle a = 1\$ , \$\displaystyle b = 3b\$ , \$\displaystyle c = -b^2 + 1\$

substitute these values into the discriminant equation ...

\$\displaystyle 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 :D
• Oct 22nd 2011, 04:58 PM
Chaim
Re: How do you solve a quadratic function when 'c' has a variable?
Quote:

Originally Posted by skeeter
\$\displaystyle a = 1\$ , \$\displaystyle b = 3b\$ , \$\displaystyle c = -b^2 + 1\$

substitute these values into the discriminant equation ...

\$\displaystyle 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.