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Math Help - How to solve for a variable in a quadratic function with 2 variables?

  1. #1
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    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.
    Last edited by Chaim; October 23rd 2011 at 06:18 PM.
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  2. #2
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    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
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  3. #3
    Super Member Quacky's Avatar
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    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.
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    Re: How to solve for a variable in a quadratic function with 2 variables?

    Quote Originally Posted by Quacky View Post
    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
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  5. #5
    Super Member Quacky's Avatar
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    Re: How to solve for a variable in a quadratic function with 2 variables?

    Quote Originally Posted by Chaim View Post
    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.
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    Re: How to solve for a variable in a quadratic function with 2 variables?

    Quote Originally Posted by Quacky View Post
    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.
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  7. #7
    Super Member Quacky's Avatar
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    Re: How to solve for a variable in a quadratic function with 2 variables?

    Quote Originally Posted by Chaim View Post
    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.
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