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.

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

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:

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

By using the vertex-form of a quadratic?

There are a few things I don't understand:

-Where has the variable $\displaystyle d$ come from?

-How have you deduced that $\displaystyle d=\frac{\pm{2}}{\sqrt{13}}$? There is no reference at all to a variable '$\displaystyle d$' in the original equation.

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

We have:

$\displaystyle a=-1$

$\displaystyle b_2=3x$

$\displaystyle c=x^2+1$

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.

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:

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

By using the vertex-form of a quadratic?

There are a few things I don't understand:

-Where has the variable $\displaystyle d$ come from?

-How have you deduced that $\displaystyle d=\frac{\pm{2}}{\sqrt{13}}$? There is no reference at all to a variable '$\displaystyle d$' in the original equation.

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

Sorry, I accidently put a d instead of a b xD

We have:

$\displaystyle a=-1$

$\displaystyle b_2=3x$

$\displaystyle c=x^2+1$

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

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:

$\displaystyle a=1$

$\displaystyle b_2=3b$

$\displaystyle c=-b^2+1$

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

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:

$\displaystyle a=1$

$\displaystyle b_2=3b$

$\displaystyle c=-b^2+1$

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

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 $\displaystyle x$, I agree.

-You can then back-substitute.

Post an attempt, see how you get on.