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Note: This is NOT question. This is igbiw rip-off tutorial. How to solve the quadratic by analyzing the roots.

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Assume quadratic \(\Huge ax^2+bx+c=0\) has roots \(\Huge \alpha\) and \(\Huge \beta\)
Then, \((x-\alpha)(x-\beta)=0\), which is an alternate formulation of our equation \(ax^2+bx+c=0\)
Now, we shall expand this alternate formulation to get \(x^2-(\alpha+\beta)x+\alpha\beta=0\) Now for our original equation we divide by a to get the equation \(\large x^2+\frac{b}{a}x+\frac{c}{a}=0\). This is also known as the monic form of a polynomial. Now, since we assume these two equations to be equivalent, we can match up the coefficients.

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In other words, \(\Large -(\alpha+\beta)=\frac{b}{a}\) and \(\Large\alpha\beta=\frac{c}{a}\)
Now, we shall do something similar to the "completing the square" method you have learned, but, this time, it will be a different method. We are seeking to express \(\large \alpha\) and \(\large \beta\) in terms of a, b, and c. Now, this is where the "magic" comes in. We will not choose to do a substitution, because the resulting equation will also be a quadratic. Instead, we shall transform these equations into an equation of the form \(\alpha-\beta=?\), in which we can easily use elimination with the first equation \(-(\alpha+\beta)=\frac{b}{a}\). For this method an understanding of the these two fundamental factoring methods is required. 1. \((a+b)^2=a^2+2ab+b^2\) 2. \((a-b)^2=a^2-2ab+b^2\) Note that for these two, a and b have nothing to do with our quadratic, I am just using them as dummy variables. These two are factoring formulas you are probably familiar with.
Back to our pair of equations. Let's call \(\Huge -(\alpha+\beta)=\frac{b}{a}\) equation 1, and \(\Huge \alpha\beta=\frac{c}{a}\) equation 2. Let's rewrite equation 1 as \(\Huge \alpha+\beta=-\frac{b}{a}\) Now for the fun part. Square equation 1 to get \(\Huge (\alpha+\beta)^2=(-\frac{b}{a})^2\) =>\(\Huge \alpha^2+2\alpha\beta+\beta^2=(-\frac{b}{a})^2\)
Subtract 4 times equation two to equation 1(shown below) \(\Huge 4*\alpha\beta=4*(\frac{c}{a})\) -> 4 times equation 2 \(\Large \alpha^2+2\alpha\beta+\beta^2-4*\alpha\beta=(-\frac{b}{a})^2-4*\frac{c}{a}\) Which simplifies down to \(\Huge \alpha^2-2\alpha\beta+\beta^2=\frac{b^2}{a^2}-4*\frac{c}{a}\) Rewrite the left side. \(\Huge (\alpha-\beta)^2=\frac{b^2}{a^2}-4*\frac{c}{a}\) Rewrite the right side. \(\Huge (\alpha-\beta)^2=\frac{b^2}{a^2}-\frac{4ac}{a^2}\) Rewrite the right side again. \(\Huge (\alpha-\beta)^2=\frac{b^2-4ac}{a^2}\) Take the plus minus square root of both sides. \(\Huge (\alpha-\beta)=\pm\sqrt{\frac{b^2-4ac}{a^2}}\) Simplify \(\Huge \alpha-\beta=\pm\frac{\sqrt{b^2-4ac}}{a}\) now, simply add the rewritten equation 1 (shown below) \(\Huge \alpha+\beta=-\frac{b}{a}\) to our new found equation (shown below) \(\Huge \alpha-\beta=\pm\frac{\sqrt{b^2-4ac}}{a}\)
So, \(\large \alpha+\beta+\alpha-\beta=\frac{-b\pm\sqrt{b^2-4ac}}{a}\) Or, \(\Huge \alpha=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) to find beta, we simply subtract equation 2 from equation 1, or, \(\large \alpha+\beta-(\alpha-\beta)=\frac{-b\pm\sqrt{b^2-4ac}}{a}\), which we also simplify to get \(\Huge \beta=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) Now, to solve the ambiguity problem. We see that for both roots we get the same result with \(\pm\), so we know that one of the roots is +, and one is -. We can just call \(\alpha\) the root with the principle square root, and \(\beta\) the root that is with the negative square root. In other words, \(\Huge \alpha=\frac{-b+\sqrt{b^2-4ac}}{2a}\) \(\Huge \beta=\frac{-b-\sqrt{b^2-4ac}}{2a}\)
And that is how you solve the quadratic by looking at the roots.
Oh come on guys, you can actually read it, this one is not difficult -.- @ParthKohli @lgbasallote @apoorvk @dpaInc @mimi_x3
now i understand why.. i always just accepted it and moved on lol, thank you for getting rid of my curiosity nice tut
@amorfide , you read it?
i wouldn't comment if i never read it lol, you should do one where you have more roots of an equation where you get alpha beta and gamma
@amorfide , you can read my tutorial on the cubic, but it doesn't use these nice greek letters lol
vieta's formula?
Probably. It's probably a obscure thing I haven't heard of. Like transmauchen.
lol, you havent heard of but you know how to it...?
Hey, I'm a genius trololol
i came up with it after trying to apply cardano's solution to the cubic to the quadratic. The first part, I mean.
Hey nice explanation:)
Thanks :)
@dpaInc , did you get to see this?
Interesting. Sadly, I can't make heads or tails of it.
Eh? what part do you not understand?
Try the whole thing. :( I mean, the logic and the steps make sense, but what was the purpose? What is this supposed to do? How does it help?
@roadjester , it's merely a solution to the quadratic by reverse analysis.
Nice work, Thanks

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