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Compassionate

  • one year ago

What is the quadratic formula and how do I use it? @Zale101

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  1. Compassionate
    • one year ago
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    Patiently waits

  2. dan815
    • one year ago
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    |dw:1444022758804:dw|

  3. dan815
    • one year ago
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    it originates from completing the square method

  4. Compassionate
    • one year ago
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    What type of expressions would I need to use this one, is it limited to binomials?

  5. dan815
    • one year ago
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    all equations of the form ax^2+bx+ c= 0 a,b,c can be any real number including 0

  6. Zale101
    • one year ago
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    The quadratic formula is what i believe a short cut way to solve for x. You can factor to solve for x, but when you have unfavorable quadratics, then the quadratic formula comes to handy. From where did it came from? Well, consider the standard quadratic equation: \(\large ax^2+bx+c=0\) if you tried to solve for x, you will come up with the quadratic formula. Let us try it! It \(\large ax^2+bx+c=0\) Divide everything by a \(\large x^2+\frac{b}{a}x+\frac{c}{a}=0\) \(\large x^2+\frac{b}{a}x+\frac{b^2}{4a}-\frac{b^2}{4a}+\frac{c}{a}=0\) \(\large (x+\frac{b}{2a})^2-\frac{b^2}{4a}+\frac{c}{a}=0\) \(\large (x+\frac{b}{2a})^2=\frac{b^2}{4a}-\frac{c}{a}\) \(\large (x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\) square root both sides \(\large x+\frac{b}{2a}=\sqrt{\frac{b^2-4ac}{4a^2}}\) \(\Large x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

  7. dan815
    • one year ago
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    a=/= 0

  8. Zale101
    • one year ago
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    This is he proof for the quadratic formula. YOU don't need to do it but you can always memorize it!

  9. Zale101
    • one year ago
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    Sorry for keeping you waiting...

  10. Compassionate
    • one year ago
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    It's okay. So what is all this talk I hear about binomials in quadratic formulas? What's the point?

  11. Zale101
    • one year ago
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    binomials in the quadratic? If you have two binomials multiplying together (x-2)(x+1), you can solve for x without needing the quadratic formula. You can simply set the two binomials to zero and solve for x. However, both methods can give you the value of x.

  12. Compassionate
    • one year ago
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    Why is it if I have (x-2)(x+1) the solutions really become x = 2, x = -1

  13. Compassionate
    • one year ago
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    \[s(t) = –g \times t^2 + vt + h\] i think this would be easier to remember.

  14. zepdrix
    • one year ago
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    When you have \(\large\rm f(x)=(x-2)(x+1)\), you're asking why this gives us "solutions" of x=2 and x=-1?

  15. Compassionate
    • one year ago
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    yes

  16. zepdrix
    • one year ago
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    By "solutions", we mean zeroes, or where the function intercepts the x-axis. This happens when y=0.\[\large\rm 0=(x-2)(x+1)\]From here you need to think back to basic arithmetic. If you have \(\large\rm 0=\color{royalblue}{A}\color{orangered}{B}\), then one of the numbers, A or B, has to be 0. That's the only way you can get 0 in multiplication. Applying this to our polynomial, it tells us that the factor (x-2) or (x+1) has to be 0.\[\large\rm 0=\color{royalblue}{(x-2)}\color{orangered}{(x+1)}\]So either \(\large\rm 0=x-2\) (adding 2 to each side gives us x=2) or \(\large\rm 0=x+1\) (subtracting 1 from each side gives us x=-1)

  17. zepdrix
    • one year ago
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    \[\large\rm 0=\color{royalblue}{(x-2)}\color{orangered}{(x+1)}\]I hope that makes some sense. We can only get zero if we have this:\[\large\rm 0=\color{royalblue}{0}\color{orangered}{(x+1)}\]or this\[\large\rm 0=\color{royalblue}{(x-2)}\color{orangered}{0}\]

  18. Compassionate
    • one year ago
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    I get it. That makes sense. So what about applying the gravity formula. Let me give you an example You throw a ball at the speed of 10 m/s standing on the ground. It reaches a maximum height of 50 feet. When will it reach this height? (Time) I know there's something with a vertex, or there's a vertex formula? But this seems so confusing, because sometimes I'll need to find the height, sometimes I need to find the speed -- can you shed some light on this?

  19. Compassionate
    • one year ago
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    Scratch that. Lets say I'm given y = 3x2 + x – 2 and I am asking to not only find the vertex, but graph it as well

  20. zepdrix
    • one year ago
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    Ummm, well you have a basic understanding of what a parabola looks like, yes? If I give you \(\large\rm f=(x)^2\), it's a bowl shape `opening upward`. If I give you \(\large\rm g=-(x)^2\), it's the same shape but `opening downward`. When we throw other stuff into the mix,\[\large\rm h=-(x-2)^2+3\]We have the exact same shape, we're only moving the "base of that bowl" or vertex as we call it. Having a coefficient on the squared term though,\[\large\rm h=-3(x-2)^2+3\]will affect the growth rate and change the shape a bit.

  21. zepdrix
    • one year ago
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    So for your example: \(\large\rm y=3x^2+x-2\) Hmm this might be a tricky one to start with if you're not familiar with `completing the square`. Maybe I can use a simpler example? :D

  22. Compassionate
    • one year ago
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    Sure. Pick an example. I understand that the quadratic formula finds both x-intercepts, usually two whole positive and negative numbers, and sometime it's just one positive or negative intercept, but say I wanted to find the vertex

  23. zepdrix
    • one year ago
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    Just a note about that factoring thing earlier, and the "solutions" that we were guaranteed. Most polynomials can not be factored nicely. Especially anything dealing with physics like your motion equation. So it's usually not very easy to turn it into something like this: \(\rm 0=(x-2)(x+1)\) That's why the Quadratic Function is amazing. It's always works. If you find that more intuitive to use rather than these weird algebraic shortcuts, then by all means. That was just a side note :) Ok back on topic. Yes I have an example.

  24. zepdrix
    • one year ago
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    I'm thinking of a function like this:\[\large\rm f(x)=\color{orangered}{x^2+8x}+5\]I highlighted the x's in orange because we want to `complete the square` on them. We would like to turn f(x) into something like this:\[\large\rm f(x)=\color{orangered}{(x+number)^2}+other~number\]Lemme use letters instead of words there..\[\large\rm f(x)=\color{orangered}{(x-h)^2}+k\]When it is in this form, we call it vertex form. And the vertex of the function is located at \(\large\rm (h,k)\).

  25. zepdrix
    • one year ago
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    For our function, Think of it like this:\[\large\rm f(x)=\color{orangered}{x^2+4x+4x}+5\]I've split the middle term into two equal parts. I want to show you the square method. It's not very practical, but it gives a lot of intuition into this weird method.

  26. zepdrix
    • one year ago
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    |dw:1444025523865:dw|I start with a square.

  27. zepdrix
    • one year ago
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    |dw:1444025568081:dw|I'm going to let this length across be x and the other part be 4.

  28. zepdrix
    • one year ago
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    Since it's a square, it has equal dimension on the other side,|dw:1444025623649:dw|

  29. zepdrix
    • one year ago
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    So now we'll fill in these smaller squares. Notice the upper left square is an \(\large\rm x~by~x\) square.

  30. zepdrix
    • one year ago
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    |dw:1444025674721:dw|And the other rectangular shapes have side lengths 4 and x.

  31. zepdrix
    • one year ago
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    So in this example, to complete the square, fill it in the rest of the way, we need that 4 by 4 box filled. So we need 16, ya?

  32. zepdrix
    • one year ago
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    |dw:1444025797639:dw|Once we have that 16, this square can be simplified down into \(\large\rm (x+4)(x+4)\) those are the side lengths. Which we would further simplify to \(\large\rm (x+4)^2\)

  33. zepdrix
    • one year ago
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    Maybe I'm going into too much weird detail here :) lol

  34. zepdrix
    • one year ago
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    So umm

  35. zepdrix
    • one year ago
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    \[\large\rm f(x)=\color{orangered}{x^2+4x+4x}+5\]We'll add 16 to complete our square because that's the number we needed.\[\large\rm f(x)=\color{orangered}{x^2+4x+4x+16}+5\]But in the land of math, we're not allowed to just add 16 willy nilly like that. It's changes the function. So, in order to keep the same function, we need to balance it. A way we can do this is by subtracting 16 at the same time.\[\large\rm f(x)=\color{orangered}{x^2+4x+4x+16}-16+5\]

  36. Compassionate
    • one year ago
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    I've seen thje example, and I kind of get it, but it's a little complex for me. I'm just familiar with the quadratic formula and the vertex = -b/2a thing. I really do appreciate your time and wanting to help me, it's just some things I don't get

  37. zepdrix
    • one year ago
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    Ok if you're more comfortable with the shortcuts, we can skip right to the meat then :) \[\large\rm f(x)=\color{orangered}{x^2+8x}+5\]\[\large\rm b=8,\qquad \frac{b}{2}=4,\qquad\left(\frac{b}{2}\right)^2=16\]So we'll add 16 to complete our square, and subtract 16 at the same time to balance it.\[\large\rm f(x)=\color{orangered}{x^2+8x+16}-16+5\]And our orange part simplifies down to: \(\large\rm \left(x+\frac{b}{2}\right)^2\) That's why I listed all three of the b parts, we'll need all of them.

  38. zepdrix
    • one year ago
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    \[\large\rm f(x)=\color{orangered}{\left(x+4\right)^2}-16+5\]

  39. Compassionate
    • one year ago
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    Question: in most quadratic problem involving motion, will I generally be asked to find the time? Say, if I'm given something like 16t^2 + 96t + 200, and I solve the vertex, that will give me the time the ball reaches it's maxiumum height, and solving for the positive x-intercept will give me the time it touches the ground. In High School Math, are we just generally finding the time of objects?

  40. zepdrix
    • one year ago
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    Good good. Yes. In motion problems those are the three features you'll spend a lot of time on. Maximum height, Starting and end point.

  41. Compassionate
    • one year ago
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    In my textbooks, we never learned about completing the square. Just basic formulas and how to find them. I haven't gotten to that part in my books. I mean, I'm familiar with factoring polynomials but not completing their squares

  42. zepdrix
    • one year ago
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    So in this example we're working on:\[\large\rm f(x)=\color{orangered}{\left(x+4\right)^2}-11\]Our vertex is located at: \(\large\rm (-4,-11)\)

  43. zepdrix
    • one year ago
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    So that would give you your base point.|dw:1444026482133:dw|

  44. zepdrix
    • one year ago
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    But knowing exactly how it grows is a little tricky to graph. That's where those "solutions" come into play, ya?|dw:1444026562732:dw|Wherever it is that they might be

  45. zepdrix
    • one year ago
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    Oh, also notice... when you put it into vertex form, you'll avoid using the quadratic function altogether. You really only want to use the quadratic function from `standard form`.

  46. zepdrix
    • one year ago
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    \[\large\rm f(x)=\left(x+4\right)^2-11\]\[\large\rm 0=\left(x+4\right)^2-11\]It's much easier to solve for x explicitly from here. Add 11 to each side,\[\large\rm 11=(x+4)^2\]Square root,\[\large\rm \pm\sqrt{11}=x+4\]Subtract 4,\[\large\rm x=-4\pm\sqrt{11}\]

  47. zepdrix
    • one year ago
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    quadratic formula* not function.. sorry typo :(

  48. Compassionate
    • one year ago
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    What's the need to complete the square, whjy can't we just plug it into the quadratic formula, find the x-intercepts, then find hte vertex?

  49. Compassionate
    • one year ago
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    I understand you're knowledgeable of the subject, but I don't know how to complete squares. You're confusing me

  50. zepdrix
    • one year ago
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    Hmm. It's pretty straight forward finding the x-coordinate of your vertex. We were given this \(\large\rm x^2+8x\) for the x's. The vertex is located at \(\large\rm x=-\frac{b}{2}\). But finding the y-coordinate is a lot trickier. We don't have a nice shortcut like the quadratic formula. We simply have to put it into vertex form. If you search hard enough, maybe you can find a weird shortcut formula that allows you to skip `completing the square`. But I'm not familiar with any such method D:

  51. Compassionate
    • one year ago
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    I apprecaite your help, but it's 1:30am. Maybe we can pickup tomorrow.

  52. zepdrix
    • one year ago
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    Haha I was thinking the same thing XD Sorry for rambling on so long!

  53. zepdrix
    • one year ago
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    I get carried away sometimes hehe

  54. Compassionate
    • one year ago
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    It's great, actually, you're enthusiastic. Never be ashamed of it. TTYL

  55. zepdrix
    • one year ago
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    \c:

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