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Is this a valid proof for the quad. form.

  1. May 31, 2009 #1
    When I was much younger (back in like middle school or freshman year in HS) I always wondered how you prove the quadratic formula, and then a couple years later I just fiddled around and came up with:

    [tex] Given: ax^2 + bx + c = 0 [/tex]

    [tex] (\alpha x + \beta)^2 + \delta = 0 [/tex]

    [tex] (\alpha x + \beta)^2 + \delta = \alpha^2 x^2 + 2 \alpha \beta x + \beta^2 + \delta = 0[/tex]

    [tex] \alpha^2 x^2 + 2 \alpha \beta x + \beta^2 + \delta = ax^2 + bx + c [/tex]

    [tex] \alpha^2 = a [/tex]

    [tex] 2 \alpha \beta = b [/tex]

    [tex] \beta^2 + \delta = c [/tex]

    so

    [tex] \alpha = \sqrt{a} [/tex]

    [tex] \beta = \frac{b}{2 \sqrt{a}} [/tex]

    [tex] \delta = c - \frac{b^2}{4a} [/tex]

    [tex] (\alpha x + \beta)^2 + \delta = 0 [/tex]

    [tex] (\alpha x + \beta)^2 = -\delta [/tex]

    [tex] (\alpha x + \beta) = \pm \sqrt{-\delta} [/tex]

    [tex] \alpha x = -\beta \pm \sqrt{-\delta} [/tex]

    [tex] x = \frac{-\beta \pm \sqrt{-\delta}}{\alpha} [/tex]

    Substituting gives

    [tex] x = \frac{-\frac{b}{2 \sqrt{a}} \pm \sqrt{-(c - \frac{b^2}{4a}})}{\sqrt{a}} [/tex]

    [tex] x = \frac{-\frac{b}{2 \sqrt{a}} \pm \sqrt{\frac{b^2}{4a} - \frac{4ac}{4a}}}{\sqrt{a}} [/tex]

    [tex] x = \frac{-\frac{b}{2 \sqrt{a}} \pm \frac{\sqrt{b^2 - 4ac}}{2\sqrt{a}}}{\sqrt{a}} [/tex]

    [tex] x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2 \sqrt{a} \sqrt{a}} [/tex]

    [tex] x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2 a} [/tex]

    I'm sure there's a much simpler way, I just never figured it out lol.
     
  2. jcsd
  3. May 31, 2009 #2

    HallsofIvy

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    I'm very impressed. For "starting from scratch" that's very good! My only criticism is that that is perhaps more complicated than necessary.

    The standard way of deriving the quadratic formula is to use "completing the square"- which is very similar to what you did.

    If [itex]ax^2+ bx+ c= 0[/itex], then [itex]x^2+ (b/a)x+ (c/a)= 0[/itex].
    [tex]x^2+ (b/a)x= -c/a[/tex].
    [tex]x^2+ (b/a)x+ (b^2/4a)= (b^2/4a)- c/a[/tex]
    [tex](x+ (b/2a))^2= (b^2/4a)- c/a= frac{b^- 4ac}{4a^2}[/tex]

    Now take the square root of each side:
    [tex]x+ (b/2a)= \pm\sqrt{\frac{b^2- 4ac}{4a^2}}= \pm\frac{\sqrt{b^2- 4ac}}{2a}[/itex]
    [tex]x= -\frac{b}{2a}\pm\frac{\sqrt{b^2- 4ac}}{2a}= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}[/tex]
     
  4. May 31, 2009 #3
    I see, I guess mine was a bit more involved and more um... damn what's that term. I swear my mind is slipping a bit lately.

    Oh well, I'll edit it in if I figure out the word. Thank you for showing me the simpler proof involving CTS
     
  5. May 31, 2009 #4

    arildno

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    Another one, based on the hallowed principle of knowing what you want to get, goes as follows:

    [tex]ax^{2}+bx+c=0[/tex]
    which implies:
    [tex]4a^{2}x^{2}+4axb+b^{2}+4ac=b^{2}\to(2ax+b)^{2}=b^{2}-4ac[/tex]
    and so on..
     
  6. May 31, 2009 #5

    arildno

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    The following is Al-Qwarizmi's (9th century) proof in the following case:
    [tex]x^{2}+bx=c[/tex]

    Now, the right-hand side can be regarded as the sum of the areas of a square with side x, and 4 rectangles with sides b/4 and x, respectively.
    (That is, we may write: [tex]x^{2}+4\frac{b}{4}x=c[/tex])

    If you arrange the rectangles along the 4 sides of the squares, you see that you get a figure that lacks 4 tiny squares of side length b/4 in order to get a square with side-length x+b/2

    Thus, we add the areas of those squares to both sides of our equation, getting:
    [tex](x+\frac{b}{2})^{2}=c+4(\frac{b}{4})^{2}[/tex]

    We may now take the square root on both sides and solve for x.
     
  7. May 31, 2009 #6
    It's much easier to start with the formula, put it in the form of a quadratic equation, then reverse your steps. You get a valid proof, but it doesn't give much insight into how you'd derive the quadratic formula in the first place.
     
  8. May 31, 2009 #7
    Guys I know this is off-topic but what is the name of that sort of proof that doesn't use assumptions or shortcuts and goes about it in the long way (eg. like my proof up in the OP). I cannot for the life of me remember the name of the term.

    Tibarn: Well the point is to prove the quadratic formula :P. We can't just directly assume it exists right at the start now can we?
     
  9. May 31, 2009 #8
    I'm not sure what you mean by no assumptions and shortcuts, but you may be thinking of "brute force".

    No, but sometimes if the formula for something is simple enough, you'll figure it out and then you can work on proving it. A common example is to discover that 1+2+...+n=n(n+1)/2 and then prove it by induction. It wouldn't be so easy to do this with the quadratic formula though.
     
  10. May 31, 2009 #9
    That was it. I can't believe I couldn't think of that. My head is all into the dark ages of the universe and cosmology and it totally shoved "brute force" into archives. Thank you!
     
  11. May 31, 2009 #10

    symbolipoint

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    I like the discussion from Arildno (barely having read it, but a quick look suggests it has the right idea). Can a good derivation be taken as equivalent to a proof? The solution for a quadratic equation is easy to derive.
     
  12. Jun 1, 2009 #11

    HallsofIvy

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    Yes, if a derivation is step by step correct, that is a proof that the formula derived is correct.
     
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