1) Derivation of the quadratic formula

This is note 11 in a set of notes showing how to obtain formulas. There will be no words beyond these short paragraphs as the rest will either consist of images or algebra showing the steps needed to derive the formula mentioned in the title.

Suggestions for other formulas to derive are welcome, however whether they are completed or not depends on my ability to derive them. The suggestions given aren't guaranteed to be the next one in the set but they will be done eventually.

1 ax2+bx+c=0\large ax^2 + bx + c = 0

2 x2+bax+ca=0\large x^2 + \frac{b}{a}x + \frac{c}{a} = 0

3.1 x2+bax=(x+b2a)2(b2a)2\large x^2 + \frac{b}{a}x = \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2

3.2 (x+b2a)2(b2a)2+ca=0\large \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} = 0

4 (x+b2a)2b24a2+4ac4a2=0\large \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{4ac}{4a^2} = 0

5 (x+b2a)2=b24ac4a2\large \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}

6 x+b2a=±b24ac4a2\large x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}

7 x+b2a=±b24ac2a\large x + \frac{b}{2a} = \frac{\pm\sqrt{b^2 - 4ac}}{2a}

8 x=b±b24ac2a\large x = \frac{- b \pm\sqrt{b^2 - 4ac}}{2a}

Note by Jack Rawlin
5 years, 5 months ago

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1 vote

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Yup! This is the derivation of the quadratic formula. Great!

Pi Han Goh - 5 years, 5 months ago

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I made a YouTube video showing how this would work in arbitrary fields not of characteristic 22. The field not having characteristic 22 is so important because this avoids any of your calculations having zero determinants; in other words, the quantities 22 and 44 will be 00 modulo 22 and our prescribed condition avoids this occurring. In such a setting, steps 6, 7 and 8 will not be valid; solutions can only exist when b24acb^2 - 4ac is a square number in the field. This indicates the severe limitations of the Fundamental Theorem of Algebra in its scope only being applied in the framework of the "real number field" and the "complex number field".

A Former Brilliant Member - 2 years, 9 months ago

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