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Homework Help: Discriminant Related Question

  1. May 28, 2008 #1
    [SOLVED] Discriminant Related Question

    Question

    [tex]x^2 - 8x - 29 = (x+a)^2 +b[/tex], where a and b are constant.

    NOTE: The equals sign should be an always equals sign, so like three lines under each other :bugeye:

    (a) Find the value of a and b

    (b)Hence, or otherwise show the roots of [tex]x^2 - 8x - 29 = 0[/tex] are [tex]c =+/- d\sqrt5[/tex] where c and d are integers.

    Attempt

    (a)[tex]x^2 - 8x - 29 = (x+a)^2 +b[/tex]

    [tex](x-4)^2 - 16 - 29 = (x+a)^2 +b[/tex]

    a = -4
    b = -45

    (b) Erm...not sure, as it says roots I will assume it means two roots so [tex]b^2 - 4ac > 0[/tex]

    64 + 116 > 0, yes this is all alright but I am not sure how to get it in the form [tex]c =+/- d\sqrt5[/tex]

    _Mayday_
     
  2. jcsd
  3. May 28, 2008 #2
    Oh that's done with \equiv

    [tex]\equiv[/tex]
     
  4. May 28, 2008 #3
    Cheers!

    Anyone got any ideas on how to solve the other question? :rofl:
     
  5. May 28, 2008 #4

    HallsofIvy

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    Science Advisor

    What are the roots of the equation x2- 8x- 29= 0? Since you have already "completed the square" that should be easy!
     
  6. May 28, 2008 #5
    By find the roots do they just mean find the possible values for x? The question is what has got me, what do they mean by find the roots? I know I should know this but, I think I have already done the 'hard' bit.
     
  7. May 28, 2008 #6
    [tex](x-4)^2 = 45[/tex]

    [tex]\sqrt{45} = +/- 3\sqrt5[/tex]
     
  8. May 28, 2008 #7
    [tex]x - 4 = +\- 3\sqrt5[/tex]

    [tex]x = 4 +/- 3\sqrt5[/tex]
     
  9. May 28, 2008 #8
    I think that is correct, thank you!
     
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