inkyvoyd
  • inkyvoyd
Note: This is NOT question. This is igbiw rip-off tutorial. How to solve the quadratic by analyzing the roots.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
inkyvoyd
  • inkyvoyd
Assume quadratic \(\Huge ax^2+bx+c=0\) has roots \(\Huge \alpha\) and \(\Huge \beta\)
inkyvoyd
  • inkyvoyd
Then, \((x-\alpha)(x-\beta)=0\), which is an alternate formulation of our equation \(ax^2+bx+c=0\)
inkyvoyd
  • inkyvoyd
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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inkyvoyd
  • inkyvoyd
In other words, \(\Large -(\alpha+\beta)=\frac{b}{a}\) and \(\Large\alpha\beta=\frac{c}{a}\)
inkyvoyd
  • inkyvoyd
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.
inkyvoyd
  • inkyvoyd
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 \(\Huge (\alpha+\beta)^2=(-\frac{b}{a})^2\) =>\(\Huge \alpha^2+2\alpha\beta+\beta^2=(-\frac{b}{a})^2\)
inkyvoyd
  • inkyvoyd
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) \(\Huge \alpha-\beta=\pm\frac{\sqrt{b^2-4ac}}{a}\)
inkyvoyd
  • inkyvoyd
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, \(\Huge \alpha=\frac{-b+\sqrt{b^2-4ac}}{2a}\) \(\Huge \beta=\frac{-b-\sqrt{b^2-4ac}}{2a}\)
inkyvoyd
  • inkyvoyd
And that is how you solve the quadratic by looking at the roots.
inkyvoyd
  • inkyvoyd
Oh come on guys, you can actually read it, this one is not difficult -.- @ParthKohli @lgbasallote @apoorvk @dpaInc @mimi_x3
amorfide
  • amorfide
now i understand why.. i always just accepted it and moved on lol, thank you for getting rid of my curiosity nice tut
inkyvoyd
  • inkyvoyd
@amorfide , you read it?
amorfide
  • amorfide
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
inkyvoyd
  • inkyvoyd
@amorfide , you can read my tutorial on the cubic, but it doesn't use these nice greek letters lol
Mimi_x3
  • Mimi_x3
vieta's formula?
inkyvoyd
  • inkyvoyd
Probably. It's probably a obscure thing I haven't heard of. Like transmauchen.
Mimi_x3
  • Mimi_x3
lol, you havent heard of but you know how to it...?
inkyvoyd
  • inkyvoyd
Hey, I'm a genius trololol
inkyvoyd
  • inkyvoyd
i came up with it after trying to apply cardano's solution to the cubic to the quadratic. The first part, I mean.
maheshmeghwal9
  • maheshmeghwal9
Hey nice explanation:)
inkyvoyd
  • inkyvoyd
Thanks :)
maheshmeghwal9
  • maheshmeghwal9
yw:)
inkyvoyd
  • inkyvoyd
@dpaInc , did you get to see this?
roadjester
  • roadjester
Interesting. Sadly, I can't make heads or tails of it.
inkyvoyd
  • inkyvoyd
Eh? what part do you not understand?
roadjester
  • roadjester
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?
inkyvoyd
  • inkyvoyd
@roadjester , it's merely a solution to the quadratic by reverse analysis.
anonymous
  • anonymous
Nice work, Thanks

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