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Note: This is NOT question. This is igbiw ripoff tutorial.
How to solve the quadratic by analyzing the roots.
 one year ago
 one year ago
Note: This is NOT question. This is igbiw ripoff tutorial. How to solve the quadratic by analyzing the roots.
 one year ago
 one year ago

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inkyvoydBest ResponseYou've already chosen the best response.20
Assume quadratic \(\Huge ax^2+bx+c=0\) has roots \(\Huge \alpha\) and \(\Huge \beta\)
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
Then, \((x\alpha)(x\beta)=0\), which is an alternate formulation of our equation \(ax^2+bx+c=0\)
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
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.
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
In other words, \(\Large (\alpha+\beta)=\frac{b}{a}\) and \(\Large\alpha\beta=\frac{c}{a}\)
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
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. \((ab)^2=a^22ab+b^2\) 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.
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
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 \(\Huge (\alpha+\beta)^2=(\frac{b}{a})^2\) =>\(\Huge \alpha^2+2\alpha\beta+\beta^2=(\frac{b}{a})^2\)
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
Subtract 4 times equation two to equation 1(shown below) \(\Huge 4*\alpha\beta=4*(\frac{c}{a})\) > 4 times equation 2 \(\Large \alpha^2+2\alpha\beta+\beta^24*\alpha\beta=(\frac{b}{a})^24*\frac{c}{a}\) Which simplifies down to \(\Huge \alpha^22\alpha\beta+\beta^2=\frac{b^2}{a^2}4*\frac{c}{a}\) 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. \(\Huge (\alpha\beta)^2=\frac{b^24ac}{a^2}\) Take the plus minus square root of both sides. \(\Huge (\alpha\beta)=\pm\sqrt{\frac{b^24ac}{a^2}}\) Simplify \(\Huge \alpha\beta=\pm\frac{\sqrt{b^24ac}}{a}\) now, simply add the rewritten equation 1 (shown below) \(\Huge \alpha+\beta=\frac{b}{a}\) to our new found equation (shown below) \(\Huge \alpha\beta=\pm\frac{\sqrt{b^24ac}}{a}\)
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
So, \(\large \alpha+\beta+\alpha\beta=\frac{b\pm\sqrt{b^24ac}}{a}\) Or, \(\Huge \alpha=\frac{b\pm\sqrt{b^24ac}}{2a}\) to find beta, we simply subtract equation 2 from equation 1, or, \(\large \alpha+\beta(\alpha\beta)=\frac{b\pm\sqrt{b^24ac}}{a}\), which we also simplify to get \(\Huge \beta=\frac{b\pm\sqrt{b^24ac}}{2a}\) 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, \(\Huge \alpha=\frac{b+\sqrt{b^24ac}}{2a}\) \(\Huge \beta=\frac{b\sqrt{b^24ac}}{2a}\)
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
And that is how you solve the quadratic by looking at the roots.
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
Oh come on guys, you can actually read it, this one is not difficult . @ParthKohli @lgbasallote @apoorvk @dpaInc @mimi_x3
 one year ago

amorfideBest ResponseYou've already chosen the best response.0
now i understand why.. i always just accepted it and moved on lol, thank you for getting rid of my curiosity nice tut
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
@amorfide , you read it?
 one year ago

amorfideBest ResponseYou've already chosen the best response.0
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
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
@amorfide , you can read my tutorial on the cubic, but it doesn't use these nice greek letters lol
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
http://openstudy.com/study#/updates/4fa231ede4b029e9dc332a4f Then http://openstudy.com/users/inkyvoyd#/updates/4fa33e9fe4b029e9dc3409ff
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
Probably. It's probably a obscure thing I haven't heard of. Like transmauchen.
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
lol, you havent heard of but you know how to it...?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
Hey, I'm a genius trololol
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
i came up with it after trying to apply cardano's solution to the cubic to the quadratic. The first part, I mean.
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.3
Hey nice explanation:)
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
@dpaInc , did you get to see this?
 one year ago

roadjesterBest ResponseYou've already chosen the best response.0
Interesting. Sadly, I can't make heads or tails of it.
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
Eh? what part do you not understand?
 one year ago

roadjesterBest ResponseYou've already chosen the best response.0
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?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.20
@roadjester , it's merely a solution to the quadratic by reverse analysis.
 one year ago

supercrazy92Best ResponseYou've already chosen the best response.0
Nice work, Thanks
 one year ago
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