Problem deriving the quadratic formula

Dale10101

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I cannot rationalize why I come up with the slightly wrong derivation, please see page 2. Have resolved this issue, thanks.
quadratic_form.jpgquadratic_form1.jpg
 
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I cannot rationalize why I come up with x =[ +/- b +/- (b^2-4ac)^(1/2)]/2a instead of simply x =[ b +/- (b^2-4ac)^(1/2)]/2a. Help appreciated, thank you,
View attachment 3166View attachment 3167

I cannot rationalize why I come up with x =[ +/- b +/- (b^2-4ac)^(1/2)]/2a instead of simply x =[ b +/- (b^2-4ac)^(1/2)]/2a. Help appreciated, thank you,
Neither is correct, not to mention that you need grouping symbols around your denominator.

It should be: x = [-b +/- (b^2 - 4ac)^(1/2)]/(2a).
 
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The formula I wrote in the question is not the error I made in the derivation.

Neither is correct, not to mention that you need grouping symbols around your denominator.

It should be: x = [-b +/- (b^2 - 4ac)^(1/2)]/(2a).

Please see page 2 of the derivation.
 
x = [-b +or- (b^2 - 4ac)^(1/2)] / (2a) \(\displaystyle \ \ \ \ \ \ \ \) The "/" is to go right up against the "]" and the "(,"
otherwise the intended numerator and denominator are just "floating" on opposite sides of the "/" symbol.



/ is divide symbol.
No, sometimes it is the "divide symbol." Check the context.


"+/-" positive or negative, as the case may be
 
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Wait, I think I see what is happening.

Correction to the third statement in my derivation.

quadratic_form3.jpgquadratic_form4.jpg
 
?

-(-) is + not +/-
Sorry, I am not sure what statement you are pointing to. I tried to number most statements, can you be more specific please?

Oh, maybe you are referring to the final statements at the end of page 2. On a later post, I explained the false case I introduced that resulted in that expression, page 3.

That last expression was intended to convey the four possible solutions that I had derived. I knew something was wrong! I mean, I would be on talk shows right now if I had proved two additional solutions provided by the quadratic formula. wah-wah-wah :( me.
 
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I've always thought that completing the square was the easiest way to derive the quadratic formula:

\(\displaystyle ax^2+bx+c=0\)

\(\displaystyle ax^2+bx=-c\)

\(\displaystyle x^2+\dfrac{b}{a}x=\dfrac{-c}{a}\)

\(\displaystyle \left (\dfrac{b}{2a} \right)^2=\dfrac{b^2}{4a^2}\)

\(\displaystyle x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}=\dfrac{-c}{a}+\dfrac{b^2}{4a^2}\)

\(\displaystyle \left (x+\dfrac{b}{2a} \right)^2=\dfrac{b^2-4ac}{4a^2}\)

\(\displaystyle x+\dfrac{b}{2a}=\pm \sqrt{\dfrac{b^2-4ac}{4a^2}}\)

\(\displaystyle x+\dfrac{b}{2a}=\dfrac{\pm \sqrt{b^2-4ac}}{2a}\)

\(\displaystyle x=\dfrac{-b}{2a} \pm \dfrac{\sqrt{b^2-4ac}}{2a}\)

\(\displaystyle x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\)
 
Thanks

Thanks for your clean derivation, yes, the square method seems most direct, hope it wasn't too much work translating to LaTex.

I've always thought that completing the square was the easiest way to derive the quadratic formula:

\(\displaystyle ax^2+bx+c=0\)

\(\displaystyle ax^2+bx=-c\)

\(\displaystyle x^2+\dfrac{b}{a}x=\dfrac{-c}{a}\)

\(\displaystyle \left (\dfrac{b}{2a} \right)^2=\dfrac{b^2}{4a^2}\)

\(\displaystyle x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}=\dfrac{-c}{a}+\dfrac{b^2}{4a^2}\)

\(\displaystyle \left (x+\dfrac{b}{2a} \right)^2=\dfrac{b^2-4ac}{4a^2}\)

\(\displaystyle x+\dfrac{b}{2a}=\pm \sqrt{\dfrac{b^2-4ac}{4a^2}}\)

\(\displaystyle x+\dfrac{b}{2a}=\dfrac{\pm \sqrt{b^2-4ac}}{2a}\)

\(\displaystyle x=\dfrac{-b}{2a} \pm \dfrac{\sqrt{b^2-4ac}}{2a}\)

\(\displaystyle x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\)
 
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