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

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    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
    Last edited by Chaim; October 18th 2011 at 06:25 PM. Reason: Added 1 more
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    Re: How do you solve a quadratic function when 'c' has a variable?

    Quote Originally Posted by Chaim View Post
    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?
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    Re: How do you solve a quadratic function when 'c' has a variable?

    Quote Originally Posted by skeeter View Post
    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.
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    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
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    Re: How do you solve a quadratic function when 'c' has a variable?

    Quote Originally Posted by skeeter View Post
    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?
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    Re: How do you solve a quadratic function when 'c' has a variable?

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

    Quote Originally Posted by skeeter View Post
    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?
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    Re: How do you solve a quadratic function when 'c' has a variable?

    Quote Originally Posted by Chaim View Post
    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.
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    Re: How do you solve a quadratic function when 'c' has a variable?

    Quote Originally Posted by skeeter View Post
    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
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    Re: How do you solve a quadratic function when 'c' has a variable?

    Quote Originally Posted by skeeter View Post
    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.
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    Re: How do you solve a quadratic function when 'c' has a variable?

    Quote Originally Posted by Chaim View Post
    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
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    Re: How do you solve a quadratic function when 'c' has a variable?

    Quote Originally Posted by skeeter View Post
    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
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    Re: How do you solve a quadratic function when 'c' has a variable?

    Quote Originally Posted by skeeter View Post
    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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