## inkyvoyd 4 years ago Note: This is NOT question. This is igbiw rip-off tutorial. How to solve the quadratic by analyzing the roots.

1. inkyvoyd

Assume quadratic $$\Huge ax^2+bx+c=0$$ has roots $$\Huge \alpha$$ and $$\Huge \beta$$

2. inkyvoyd

Then, $$(x-\alpha)(x-\beta)=0$$, which is an alternate formulation of our equation $$ax^2+bx+c=0$$

3. 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.

4. inkyvoyd

In other words, $$\Large -(\alpha+\beta)=\frac{b}{a}$$ and $$\Large\alpha\beta=\frac{c}{a}$$

5. 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.

6. 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 $$\Huge (\alpha+\beta)^2=(-\frac{b}{a})^2$$ =>$$\Huge \alpha^2+2\alpha\beta+\beta^2=(-\frac{b}{a})^2$$

7. 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) $$\Huge \alpha-\beta=\pm\frac{\sqrt{b^2-4ac}}{a}$$

8. 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, $$\Huge \alpha=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ $$\Huge \beta=\frac{-b-\sqrt{b^2-4ac}}{2a}$$

9. inkyvoyd

And that is how you solve the quadratic by looking at the roots.

10. inkyvoyd

Oh come on guys, you can actually read it, this one is not difficult -.- @ParthKohli @lgbasallote @apoorvk @dpaInc @mimi_x3

11. amorfide

now i understand why.. i always just accepted it and moved on lol, thank you for getting rid of my curiosity nice tut

12. inkyvoyd

13. 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

14. inkyvoyd

@amorfide , you can read my tutorial on the cubic, but it doesn't use these nice greek letters lol

15. inkyvoyd
16. Mimi_x3

vieta's formula?

17. inkyvoyd

Probably. It's probably a obscure thing I haven't heard of. Like transmauchen.

18. Mimi_x3

lol, you havent heard of but you know how to it...?

19. inkyvoyd

Hey, I'm a genius trololol

20. inkyvoyd

i came up with it after trying to apply cardano's solution to the cubic to the quadratic. The first part, I mean.

21. maheshmeghwal9

Hey nice explanation:)

22. inkyvoyd

Thanks :)

23. maheshmeghwal9

yw:)

24. inkyvoyd

@dpaInc , did you get to see this?

26. inkyvoyd

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?

28. inkyvoyd

@roadjester , it's merely a solution to the quadratic by reverse analysis.

29. anonymous

Nice work, Thanks