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anonymous

  • 5 years ago

How do you solve the cubic function x^3-14x^2+16x+12=0 without using a calculator?

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  1. amistre64
    • 5 years ago
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    can do trial and error :)

  2. dumbcow
    • 5 years ago
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    first try factor by grouping

  3. amistre64
    • 5 years ago
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    lets make a pool of useable numbers that will narrow our solutions: factors of last# +- 1,12,2,6,3,4 -------------: ------------- factors of first# +- 1

  4. amistre64
    • 5 years ago
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    descartes has a sign theorum; the number of sign changes tells us the number of + solutiuons available: possible 2 (+) solutions in this case

  5. amistre64
    • 5 years ago
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    the next best thing I can do with this is to start trying and erroring :)

  6. amistre64
    • 5 years ago
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    x=1; lets try that.. 1| 1 -14 16 12 0 1 -13 3 --------------- 1 -13 3 15 <-- not zero in the end; not a solution

  7. amistre64
    • 5 years ago
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    x=2; lets try that.. 2| 1 -14 16 12 0 2 -24 16 --------------- 1 -12 8 28 <-- not a zero, not a solution...

  8. anonymous
    • 5 years ago
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    I feel like this could take days to actually figure out haha...what if the answer is not a whole number?

  9. amistre64
    • 5 years ago
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    x = 6 maye?? 6| 1 -14 16 12 0 6 -48 ?? <-- not make it zero; not a solution --------------- 1 -8 -36

  10. amistre64
    • 5 years ago
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    lol..there has to be at least one solution. All degrees higher than a 2 can be factored to at least a linear and a irreducible quadratic

  11. amistre64
    • 5 years ago
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    and one of our pool options has got to work :)

  12. amistre64
    • 5 years ago
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    x=3; lets try that.. 3| 1 -14 16 12 0 3 -33 ........ not a -12 that we need here --------------- 1 -11-27 ..........not the zero we need here

  13. amistre64
    • 5 years ago
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    -1? x=2; lets try that.. -1| 1 -14 16 12 0 -1 15 -21 --------------- 1 -15 21 ..... <-- not a zero, not a solution

  14. amistre64
    • 5 years ago
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    copy paste edit and forgot to clean that x=2 outta there lol

  15. amistre64
    • 5 years ago
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    x=-2; lets try that.. -2| 1 -14 16 12 0 -2 32 -56 --------------- 1 -16 48 ..... <-- not a zero, not a solution

  16. anonymous
    • 5 years ago
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    No doubt there are solutions for x but is it possible for all 3 solutions for x to be a fraction? If that's the case, then guessing whole numbers will lead to infinite frustration ha.

  17. amistre64
    • 5 years ago
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    4? 4| 1 -14 16 12 0 4 -40 -96 --------------- 1 -10 -24 28 <-- not a zero, not a solution well, since our first# is a 1..that makes our pool limited to integers only; becasue they all have a 1 on the bottom :)

  18. amistre64
    • 5 years ago
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    -4? -4| 1 -14 16 12 0 -4 72 -342 --------------- 1 -18 88 ....... <-- not a zero, not a solution

  19. anonymous
    • 5 years ago
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    try 800189074059/1562500000000

  20. dumbcow
    • 5 years ago
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    there are no integer factors

  21. anonymous
    • 5 years ago
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    I think I must have made a mistake in coming up with this cubic function because I don't see how my textbook would expect me to come up with that as an option ha

  22. amistre64
    • 5 years ago
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    HOW YOU FIND THAT WITHOUT A CALCULATor...caps got locked .....too lazy to erase and re type lol

  23. dumbcow
    • 5 years ago
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    if you know calculus, newtons method works pretty well at approximating zeroes

  24. dumbcow
    • 5 years ago
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    no i cheated and used calculator :)

  25. anonymous
    • 5 years ago
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    I cheated and used a calculator as well haha

  26. amistre64
    • 5 years ago
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    i was using one after the numbers started to all look the same ;)

  27. amistre64
    • 5 years ago
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    why do the normal rules not apply to this equation?

  28. anonymous
    • 5 years ago
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    The original problem was solve the system: y1'=4y_1+2y_2+2y_3 y2'=2y_1+4y_2+2y_3 y3'=2y_1+2y_2+4y_3

  29. amistre64
    • 5 years ago
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    lol..did you give me a broken quesition :)

  30. anonymous
    • 5 years ago
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    and I used the characteristic eqution |lambda I - A|=0 and by using that I cam up with that cubic function and I didn't know how to solve it

  31. anonymous
    • 5 years ago
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    I didn't purposely give you a broken problem if I did haha...I'm just as confused as you. I just came up with that function after taking the determinate of: [(x-4), -2, -2; -2, (x-4), -2; -2, -2, (x-4)] and equalling it to zero.

  32. amistre64
    • 5 years ago
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    maaa!!! lina broke the math again ;)

  33. amistre64
    • 5 years ago
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    i wonder if its possible to decompose a cubic ...

  34. anonymous
    • 5 years ago
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    I'm sure there is somehow...ya, I think I must up the algebra of that problem. I resolved and came up with x^3-12x^2+36x-32=0 and I think 2 works for this one...sorry about the crazy cubic. ha

  35. dumbcow
    • 5 years ago
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    hmm i think i figured it out x = 2 and 8 if you dont expand everything when doing the determinant and leave it as (x-4) it makes it easier say u = x-4 then u^3-12u-16 =0

  36. amistre64
    • 5 years ago
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    yeah, 2 and 8 are roots to that one :)

  37. anonymous
    • 5 years ago
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    Ya, that's the correct answer. I just did what amistre suggested to solve the "broken one" I gave him ha. I guessed 2 and it worked. Then I had to do long division to figure out that (x^2-10x+16) was the other factor.

  38. amistre64
    • 5 years ago
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    <html> <body> <input type=button value="start" onclick="qwe()"> </body> <script language=javascript> var y,x,j,m,n function qwe() { for (x=0;x<33;x=x+1) {y = (x*x*x)+(-12*x*x)+(36*x)-32 if (y==0){alert(x+" is a root")} } alert("no root") } </script> </html>

  39. amistre64
    • 5 years ago
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  40. amistre64
    • 5 years ago
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    i should put an input box to evaluate any function entered lol

  41. dumbcow
    • 5 years ago
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    well yes thats the smart way to do it

  42. amistre64
    • 5 years ago
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    its hard to make these in the dark :)

  43. anonymous
    • 5 years ago
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    amistre , thats wrong , a polynomial of degree greater than 2 doesnt have to factor into a linear and irreducible quadratic. you can have say x^4 + 1 = 0 . no linear factor

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