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inkyvoyd Group Title

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

  • 2 years ago
  • 2 years ago

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  1. inkyvoyd Group Title
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    Assume quadratic \(\Huge ax^2+bx+c=0\) has roots \(\Huge \alpha\) and \(\Huge \beta\)

    • 2 years ago
  2. inkyvoyd Group Title
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    Then, \((x-\alpha)(x-\beta)=0\), which is an alternate formulation of our equation \(ax^2+bx+c=0\)

    • 2 years ago
  3. inkyvoyd Group Title
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    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.

    • 2 years ago
  4. inkyvoyd Group Title
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    In other words, \(\Large -(\alpha+\beta)=\frac{b}{a}\) and \(\Large\alpha\beta=\frac{c}{a}\)

    • 2 years ago
  5. inkyvoyd Group Title
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    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.

    • 2 years ago
  6. inkyvoyd Group Title
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    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 \(\Huge (\alpha+\beta)^2=(-\frac{b}{a})^2\) =>\(\Huge \alpha^2+2\alpha\beta+\beta^2=(-\frac{b}{a})^2\)

    • 2 years ago
  7. inkyvoyd Group Title
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    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) \(\Huge \alpha-\beta=\pm\frac{\sqrt{b^2-4ac}}{a}\)

    • 2 years ago
  8. inkyvoyd Group Title
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    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, \(\Huge \alpha=\frac{-b+\sqrt{b^2-4ac}}{2a}\) \(\Huge \beta=\frac{-b-\sqrt{b^2-4ac}}{2a}\)

    • 2 years ago
  9. inkyvoyd Group Title
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    And that is how you solve the quadratic by looking at the roots.

    • 2 years ago
  10. inkyvoyd Group Title
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    Oh come on guys, you can actually read it, this one is not difficult -.- @ParthKohli @lgbasallote @apoorvk @dpaInc @mimi_x3

    • 2 years ago
  11. amorfide Group Title
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    now i understand why.. i always just accepted it and moved on lol, thank you for getting rid of my curiosity nice tut

    • 2 years ago
  12. inkyvoyd Group Title
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    @amorfide , you read it?

    • 2 years ago
  13. amorfide Group Title
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    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

    • 2 years ago
  14. inkyvoyd Group Title
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    @amorfide , you can read my tutorial on the cubic, but it doesn't use these nice greek letters lol

    • 2 years ago
  15. Mimi_x3 Group Title
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    vieta's formula?

    • 2 years ago
  16. inkyvoyd Group Title
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    Probably. It's probably a obscure thing I haven't heard of. Like transmauchen.

    • 2 years ago
  17. Mimi_x3 Group Title
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    lol, you havent heard of but you know how to it...?

    • 2 years ago
  18. inkyvoyd Group Title
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    Hey, I'm a genius trololol

    • 2 years ago
  19. inkyvoyd Group Title
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    i came up with it after trying to apply cardano's solution to the cubic to the quadratic. The first part, I mean.

    • 2 years ago
  20. maheshmeghwal9 Group Title
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    Hey nice explanation:)

    • 2 years ago
  21. inkyvoyd Group Title
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    Thanks :)

    • 2 years ago
  22. maheshmeghwal9 Group Title
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    yw:)

    • 2 years ago
  23. inkyvoyd Group Title
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    @dpaInc , did you get to see this?

    • 2 years ago
  24. roadjester Group Title
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    Interesting. Sadly, I can't make heads or tails of it.

    • 2 years ago
  25. inkyvoyd Group Title
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    Eh? what part do you not understand?

    • 2 years ago
  26. roadjester Group Title
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    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?

    • 2 years ago
  27. inkyvoyd Group Title
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    @roadjester , it's merely a solution to the quadratic by reverse analysis.

    • 2 years ago
  28. supercrazy92 Group Title
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    Nice work, Thanks

    • 2 years ago
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