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poopsiedoodle

  • one year ago

How might one go about doing this problem? Solve for x: x^2 + 24x + 90 = 0

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  1. nooce
    • one year ago
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    Factor first :)

  2. nooce
    • one year ago
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    What factors of 90, when added together, give you 24?

  3. poopsiedoodle
    • one year ago
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    Gimme a second.

  4. nooce
    • one year ago
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    Take your time (:

  5. poopsiedoodle
    • one year ago
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    I don't think any do. 18 and 5 is the closest you can get to 24, which is 23.

  6. nooce
    • one year ago
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    Hmm, hold on a sec

  7. poopsiedoodle
    • one year ago
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    I will do that.

  8. nooce
    • one year ago
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    Well, since it isn't factorable, we have to use the quadratic formula. Have you learned it?

  9. poopsiedoodle
    • one year ago
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    Hardly. I've just started on it.

  10. nooce
    • one year ago
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    Well, this may take some explaining to do, so hold on for a bit haha.

  11. nooce
    • one year ago
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    x^2 + 24x + 90 = 0 Let's break down what this problem means. By plugging in a certain number(s) for x, you get zero. It's pretty much like plotting a graph of x^2 + 24x + 90 and looking for where the graph touches zero. Did I lose you yet?

  12. nooce
    • one year ago
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    |dw:1358120873885:dw|

  13. poopsiedoodle
    • one year ago
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    Okay. I get it so far.

  14. nooce
    • one year ago
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    Alright. So since we have an equation that isn't factorable, we find the zeroes by plugging it into the quadratic formula.

  15. nooce
    • one year ago
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    This is the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac} }{ 2a }\] Now, a, b, and c are the numbers in your equation -- \[a^2+bx+c\] In this case: \[x^2 + 24x + 90\] so, a=1, b= 24, and c=90.

  16. poopsiedoodle
    • one year ago
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    Ok, lemme try it now.

  17. nooce
    • one year ago
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    Remember, the ± symbol indicates that there are two expressions! (ex. 1 ± 3 is 1 + 3 AND 1 - 3)

  18. poopsiedoodle
    • one year ago
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    \[\Large0= \frac{ -24 \pm \sqrt{24^{2} -4(90)} }{ 2 }\]

  19. nooce
    • one year ago
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    Good so far!

  20. nooce
    • one year ago
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    except, it's x =

  21. poopsiedoodle
    • one year ago
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    Well, since we're already using x in the original problem, how about Y?

  22. nooce
    • one year ago
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    But we're trying to solve for x, haha

  23. poopsiedoodle
    • one year ago
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    eh, whatever. By the way, I'm having a slight problem with \[\sqrt{4(90)}\]

  24. nooce
    • one year ago
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    That's just \[\sqrt{4 • 90} = \sqrt{360}\]

  25. poopsiedoodle
    • one year ago
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    Ah. \[\Large x= \frac{ -24 \pm 24\sqrt{360} }{ 2 }\]

  26. nooce
    • one year ago
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    No, remember 24^2 is under the radical (Don't take it out!) \[\sqrt{24^2 - 360}\]

  27. poopsiedoodle
    • one year ago
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    \[\Large x= \frac{ -24 \pm \sqrt{ 24^{2} (-360)} }{ 2 }\]*

  28. poopsiedoodle
    • one year ago
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    But, since √24^2 = 24, what's wrong with taking it out?

  29. nooce
    • one year ago
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    Because we can't have a negative number under a square root, haha

  30. poopsiedoodle
    • one year ago
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    Ah.

  31. nooce
    • one year ago
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    But anyways, after plugging all of those numbers in, what do you geT?

  32. poopsiedoodle
    • one year ago
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    So, \[\Large x= { -12 \pm (\sqrt{ 24^{2} (-360)} /2)}\] ? I'm confused now e_o

  33. nooce
    • one year ago
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    Well, the /2 should be on the bottom of the whole thing, but besides that, Let's simplify the square root first. What is 24^2 - 360?

  34. poopsiedoodle
    • one year ago
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    216.

  35. poopsiedoodle
    • one year ago
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    and the root of that is 14.giganticdecimal

  36. nooce
    • one year ago
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    Well, let's simplify it haha sqrt of 216 can be reduced into what? (hint: 36 x 6 = 216)

  37. poopsiedoodle
    • one year ago
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    oh gee, this is a tought one... 2... and... erm... GIMME A MINUTE

  38. poopsiedoodle
    • one year ago
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    tough*

  39. poopsiedoodle
    • one year ago
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    OOH! I KNOW! 36 AND 6. BOOM. Don't you wish you had thought of that?

  40. nooce
    • one year ago
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    Hahahaha. You are correct x)

  41. nooce
    • one year ago
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    So... now we have \[x = \frac{ -24 \pm 6\sqrt{6} }{ 2 }\] which breaks down to: \[x = \frac{ -24 + 6\sqrt{6} }{ 2 }\] \[x = \frac{ -24 - 6\sqrt{6} }{ 2 }\] I assume you can solve from there :)

  42. poopsiedoodle
    • one year ago
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    \[\Huge x = { -12 - 3\sqrt{3} }\] ?

  43. nooce
    • one year ago
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    Annnnnnnnnd?

  44. poopsiedoodle
    • one year ago
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    I was wondering if I was right, which I guess I am. So, is it \(\Huge x=-12 -3?\)

  45. nooce
    • one year ago
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    Wait, no don't take out the radical 3 lol You need to solve for the other part: \[x = \frac{ -24 + 6\sqrt{6} }{ 2 }\] Your first solution was correct, btw!

  46. poopsiedoodle
    • one year ago
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    Oh. x=-12 + 3√3. And now what do I do?

  47. nooce
    • one year ago
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    Well, those are your solutions, haha x=-12 + 3√3 and x=-12 - 3√3 Yay!

  48. poopsiedoodle
    • one year ago
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    :D

  49. nooce
    • one year ago
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    :D indeed! If you need more help, just pm me (:

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