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Finding Minimum from a completion of a square

  1. Oct 31, 2005 #1
    so the question is:

    Complete the square of x^2 + 4x - 1 and hence find the position and value of its minimum.

    AFter completing the square i have : (x+2)^2 - 5 ,

    How do I find the minimum value from that?

    Thanks in advance guys!
     
  2. jcsd
  3. Oct 31, 2005 #2

    Tide

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    What is the smallest possible value for [itex](x+2)^2[/itex]?
     
  4. Oct 31, 2005 #3
    umm so would x= -2 and y = -5

    ??
     
  5. Oct 31, 2005 #4

    TD

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    Indeed :smile:
     
  6. Oct 31, 2005 #5
    okaay another questoin....
    Factorise and hence solve the equation:

    2x^3 + 3x^2 - 8x - 12 = 0

    I can't seem to remember how to factorize this because there's a cubic expression! how do i start?
     
  7. Oct 31, 2005 #6

    TD

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    Divisors of the constant term (i.e. -12 here) are possible zeroes of your equation. Try to find such a zero and use the fact that if x = a is a zero, you can factor out (x-a).
     
  8. Oct 31, 2005 #7
    okay thanks, i'm going to try that now, by the way is that what you call the remainder theorem?
     
  9. Oct 31, 2005 #8
    Okay, so my answer was:

    (x-2)(x+2)(2x+3)

    is that right?
     
  10. Oct 31, 2005 #9

    Tide

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    You can expand that product and see if it produces your original expression.
     
  11. Oct 31, 2005 #10
    So there's this question to factorize:

    x^4 - 3x^2 - 10 = 0

    How do I start this one because I can't find a factor to go into it.
     
  12. Oct 31, 2005 #11

    TD

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    Yes, which is easy to see if you follow Tide's advice.

    Would it be easier for you to see if you let [itex]x^2 = t[/itex] so it becomes quadratic?

    That would give [tex]t^2-3t-10=0[/tex] to factor and then substitute t by x² again.
     
  13. Oct 31, 2005 #12
    so the ans is [tex]\sqrt{5}[/tex] ?
     
    Last edited: Oct 31, 2005
  14. Oct 31, 2005 #13

    Tide

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    That's one of them! :)
     
  15. Oct 31, 2005 #14
    Ohh is it +/- Square root 5?! lol
     
  16. Oct 31, 2005 #15

    TD

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    Indeed, unless you're working complex since then a quartic equation has 4 solutions. The ones you found are the only real ones though.
     
  17. Oct 31, 2005 #16
    help with this last one

    Express the function [tex]\sqrt{3}\sin{2t} - 3\cos{2t}[/tex] in the form A[tex]\sin{(2t+\alpha)[/tex]


    i have no idea what to do
     
    Last edited: Oct 31, 2005
  18. Oct 31, 2005 #17

    Tide

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    Now you're just guessing!

    [tex]t^2 - 3t - 10 = (t-5)(t+2)[/tex]
     
  19. Oct 31, 2005 #18

    TD

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    Yes, but t was x².
     
  20. Oct 31, 2005 #19
    well if you sub the [tex]x^2[/tex] back in to the equation i get [tex]\pm\sqrt{5}[/tex]. is that not correct? and what of the other question? how should i start what am i suppose to do?
     
  21. Oct 31, 2005 #20

    TD

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    I would 'expand' sin(2t+a) with the formula for addition of angles, being:

    [tex]\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b[/tex]
     
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