# How to solve for a variable in a quadratic function with 2 variables?

• Oct 23rd 2011, 04:49 PM
Chaim
How to solve for a variable in a quadratic function with 2 variables?
The problem is f(x)=x^2+3bx-b^2+1
I'm suppose to get the value of 'b'

What I started off was with the vertex formula which was -b/2a
So this is what I think is a, b, and c
a = x^2
b=3bx
c=-b^2+1

So I did -3bx/2(1) which is -3bx/2
Then I plugged the equation back in
But when I looked at the answer book, it gave me a completely different answer which was b = +- 2/square root of 13
(Plus or minus two divdided by the square root of 13)

Can anyone help me with the steps and see if my a, b, and c variables are correct?
And please do it the way that my teacher wants it and not the discriminant formula :)
Just by finding the variable, plugging it back into the equation, and simplifying it to get the value of 'b'
So what my teacher started off with was the vertex formula, but then after that, I got lost cause I did not know if my 'a, b, and c variables were right cause there is a d and an x.
• Oct 23rd 2011, 05:40 PM
Chaim
Re: How to solve for a variable in a quadratic function with 2 variables?
Would it be
a=x^2
b=3b
c=b^2+1
• Oct 23rd 2011, 05:57 PM
Quacky
Re: How to solve for a variable in a quadratic function with 2 variables?
I'm not sure I understand completely, but as I understand it:

You are trying to solve this equation for b:

$-b^2+(3x)b+(x^2+1)=0$

By using the vertex-form of a quadratic?

There are a few things I don't understand:
-Where has the variable $d$ come from?
-How have you deduced that $d=\frac{\pm{2}}{\sqrt{13}}$? There is no reference at all to a variable ' $d$' in the original equation.

If I had to *guess* I'd say that your "values" for a,b,c refer to the form $ax^2+bx+c$ of a quadratic, although you shouldn't be using " $b$" here to mean two different things so I'm going to call it $b_2$

We have:
$a=-1$
$b_2=3x$
$c=x^2+1$

because in this equation, b is the variable we are trying to solve for, not x. I don't understand at all which method you are trying to communicate, but hopefully this start will help.
• Oct 23rd 2011, 06:19 PM
Chaim
Re: How to solve for a variable in a quadratic function with 2 variables?
Quote:

Originally Posted by Quacky
I'm not sure I understand completely, but as I understand it:

You are trying to solve this equation for b:

$-b^2+(3x)b+(x^2+1)=0$

By using the vertex-form of a quadratic?

There are a few things I don't understand:
-Where has the variable $d$ come from?
-How have you deduced that $d=\frac{\pm{2}}{\sqrt{13}}$? There is no reference at all to a variable ' $d$' in the original equation.

If I had to *guess* I'd say that your "values" for a,b,c refer to the form $ax^2+bx+c$ of a quadratic, although you shouldn't be using " $b$" here to mean two different things so I'm going to call it $b_2$
Sorry, I accidently put a d instead of a b xD
We have:
$a=-1$
$b_2=3x$
$c=x^2+1$

because in this equation, b is the variable we are trying to solve for, not x. I don't understand at all which method you are trying to communicate, but hopefully this start will help.

Sorry, I accidently put 'd' instead of a 'b' xD
But also, wouldn't I do a = -1, b = 3b, c = b^2+1?
Since after I plug it into the vertex formula: -b/2a = -3b/2(-1) = 3b/2
Since after I plug it back into the equation (subtituting), then I could find b from there, because after I plug it back in, it would be like (b)^2 +3b(b) - (b)^2 + 1 = 0
• Oct 23rd 2011, 06:36 PM
Quacky
Re: How to solve for a variable in a quadratic function with 2 variables?
Quote:

Originally Posted by Chaim
Sorry, I accidently put 'd' instead of a 'b' xD
But also, wouldn't I do a = -1, b = 3b, c = b^2+1?
Since after I plug it into the vertex formula: -b/2a = -3b/2(-1) = 3b/2
Since after I plug it back into the equation (subtituting), then I could find b from there, because after I plug it back in, it would be like (b)^2 +3b(b) - (b)^2 + 1 = 0

I can't answer this precisely without knowing your method. You're going to have to be more specific. You could, if you wanted, treat x as the main variable and b as a constant, in which case:

$a=1$
$b_2=3b$
$c=-b^2+1$

Remember that b and x are not interchangeable. $b\neq{x}$ necessarily.

But you have posted an abstract method with almost no explanation: you cannot expect us to give you a foolproof evaluation of your work when we have no idea as to what process you are trying to follow.
• Oct 23rd 2011, 06:38 PM
Chaim
Re: How to solve for a variable in a quadratic function with 2 variables?
Quote:

Originally Posted by Quacky
I can't answer this precisely without knowing your method. You're going to have to be more specific. You could, if you wanted, treat x as the main variable and b as a constant, in which case:

$a=1$
$b_2=3b$
$c=-b^2+1$

Remember that b and x are not interchangeable. $b\neq{x}$ necessarily.

But you have posted an abstract method with almost no explanation: you cannot expect us to give you a foolproof evaluation of your work when we have no idea as to what process you are trying to follow.

Well what I was thinking is that you would use the vertex formula: x=-b/2a
After you find what x equals, then you plug it back into the equation
Then solve through there, by adding, simplifying, etc.
• Oct 23rd 2011, 06:41 PM
Quacky
Re: How to solve for a variable in a quadratic function with 2 variables?
Quote:

Originally Posted by Chaim
Well what I was thinking is that you would use the vertex formula: x=-b/2a
After you find what x equals, then you plug it back into the equation
Then solve through there, by adding, simplifying, etc.

Well I'm sorry to say that I'm unfamiliar with this method. Using what I said in my previous post, do you have enough information to start?

-You can certainly put it into the vertex formula.
-By doing this, you can solve for $x$, I agree.
-You can then back-substitute.

Post an attempt, see how you get on.