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Math Help - Rearranging the quadratic formula

  1. #1
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    Rearranging the quadratic formula

    Hi guys, I've been tasked with rearranging the quadratic root formula to make "a" the subject. I've been struggling on this one for a little while now and can't figure out how to isolate a. For reference, the equation i'm referring to is the addition half of the formula, i.e.

    x = (-b + sqroot(b^2 - 4ac))/2a

    Any help is much appreciated.

    Regards,
    Sathya
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  2. #2
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    Re: Rearranging the quadratic formula

    Because that formula is a root of the equation we know that ax^2+bx+c=0 So ax^2=-bx-c So a=(-bx-c)/x^2
    Thanks from SSmith
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  3. #3
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    Re: Rearranging the quadratic formula

    Thanks very much, I thought it work be a little more complex than that to be honest, but thanks for showing me

    Very speedy response, cheers.
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  4. #4
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    Re: Rearranging the quadratic formula

    Hello, SSmith!

    biffboy has an excellent solution.
    I'll "do" the algebra.


    I've been tasked with rearranging the quadratic root formula to make a the subject.

    . . x \:=\: \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\quad a \ne 0

    We have: . x \:=\:\frac{-b \pm\sqrt{b^2-3ac}}{2a}

    Mulitply by 2a\!:\;\;2ax \;=\;-b \pm\sqrt{b^2-4ac}

    Add b\!:\;\;2ax + b \;=\;\sqrt{b^2-4ac}

    Square: . (2ax + b)^2 \;=\;\left(\sqrt{b^2-4ac}\right)^2

    . . 4a^2x^2 + 4abx + b^2 \;=\;b^2 - 4ac

    . 4a^2x^2 + 4abx + 4ac \;=\;0

    . . . 4a(ax^2 + bx + c) \;=\;0

    And we have two equations to solve.


    4a \:=\:0 \quad\Rightarrow\quad a \:=\:0\quad\text{ not allowed!}

    \underbrace{ax^2 + bx + c \:=\:0} \quad\Rightarrow\quad ax^2 \:=\:-bx - c \quad\Rightarrow\quad a \:=\:\frac{-bx-c}{x^2}

    biffboy started here!
    Thanks from SSmith and topsquark
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  5. #5
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    Re: Rearranging the quadratic formula

    Thank you so much for that extra clarification, that really helps me to understand how to get back to standard form, much appreciated Soroban!
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  6. #6
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    Re: Rearranging the quadratic formula

    this is the only confirmation that i can find on the web about subject "a" in the quadratic formula! thanks!
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