A community for students.
Here's the question you clicked on:
 0 viewing
inkyvoyd
 4 years ago
Note: This is NOT question. This is igbiw ripoff tutorial.
How to solve the quadratic by analyzing the roots.
inkyvoyd
 4 years ago
Note: This is NOT question. This is igbiw ripoff tutorial. How to solve the quadratic by analyzing the roots.

This Question is Closed

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

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

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.20Now, 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.

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

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.20Now, 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.

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.20Back 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
 4 years ago
Best ResponseYou've already chosen the best response.20Subtract 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) \(\Huge \alpha\beta=\pm\frac{\sqrt{b^24ac}}{a}\)

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.20So, \(\large \alpha+\beta+\alpha\beta=\frac{b\pm\sqrt{b^24ac}}{a}\) 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, \(\Huge \alpha=\frac{b+\sqrt{b^24ac}}{2a}\) \(\Huge \beta=\frac{b\sqrt{b^24ac}}{2a}\)

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.20And that is how you solve the quadratic by looking at the roots.

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

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

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.20@amorfide , you read it?

amorfide
 4 years ago
Best ResponseYou've already chosen the best response.0i 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
 4 years ago
Best 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

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

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.20Probably. It's probably a obscure thing I haven't heard of. Like transmauchen.

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0lol, you havent heard of but you know how to it...?

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.20Hey, I'm a genius trololol

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

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.3Hey nice explanation:)

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.20@dpaInc , did you get to see this?

roadjester
 4 years ago
Best ResponseYou've already chosen the best response.0Interesting. Sadly, I can't make heads or tails of it.

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.20Eh? what part do you not understand?

roadjester
 4 years ago
Best ResponseYou've already chosen the best response.0Try 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
 3 years ago
Best ResponseYou've already chosen the best response.20@roadjester , it's merely a solution to the quadratic by reverse analysis.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.