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Quadratic - Completing the square

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Need to solve
    2x² + 5x + 3 = 0

    2. Relevant equations

    at² + bt + c = 0
    so a = 2, b = 5, c = 3

    3. The attempt at a solution

    a x c = 6
    a + c = 5
    so ned to find somthing that times to make 6 and add to make 5. That is 2, 3

    2x² + 2x + 3x + 3

    2x (x + 1) + 3(x + 1)

    (2x + 3)(x + 1) = 0
    so x is either

    x = 3/2 = -1.5 or x = -1

    Are those answers correct?

    If so how would I go about solving this question

    A solution to the equation [tex]x^{5} - 4x^{3} = 455[/tex] lies between x = 3.6 and x = 3.8. Find the solution to 2 decimal places.

    So should change it = 0 by -455 so i get

    [tex]x^{5} - 4x^{3} -455 = 0[/tex]

    now where do i go?

    Thx
     
  2. jcsd
  3. Feb 12, 2007 #2
    Your answers are correct. However, your thread title says "completing the square" - you've solved by factoring (which is easier for this particular problem.
    Personally, I teach completing the square this way:
    2x^2 + 5x + 3 = 0
    Get rid of the 3; it's just in the way.
    2x^2 + 5x = -3
    now factor out any coefficients on the x^2.
    2(x^2 + 5/2x ) = -3
    Then, add (inside the parenthesis) half of the middle term squared.
    Since you have to add the same amount to both sides, add to the right side as well; note: whatever you add inside the parenthesis will be multiplied by 2. Thus:
    2(x^2 +5/2 x + 25/16)= -3 + 25/8
    Factor (it'll always be half the middle term again)
    2(x+5/4)^2 = 1/8
    divide by 2
    (x+5/4)^2 = 1/16
    take the +/- square root
    x+5/4 = +/- 1/4
    subtract 5/4, and you have your same answers.

    For your second problem, there are several ways to solve it, to get it correct to 2 decimal places... what methods have you learned? Show me how you would apply one of those methods and I or someone else can show you where you're making a mistake (or let you know if you're on the right track.)
     
  4. Feb 12, 2007 #3
    [tex]x^{5} - 4x^{3} -455 = 0[/tex]
    [tex]x^{5} - 4x^{3} -455 =(x^{2} - 4x)^{3} - 64 - 455 [/tex]

    is that right?
     
  5. Feb 12, 2007 #4

    HallsofIvy

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    I have no idea what you are doing. But (x2)3 is x6 not x5 so that can't be right!
     
  6. Feb 13, 2007 #5
    I think ive got it

    x³(x²-4) can be factorised even fruther to
    x³(x+2)(x-2)

    that means the equation can be x = -2, x = 2 or x=0
     
  7. Feb 13, 2007 #6

    cristo

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    This would be true if the equation was [tex]x^{5} - 4x^{3} = 0[/tex], but your equation is [tex]x^{5} - 4x^{3} = 455[/tex]. Since you're looking for an approximate value of the roots, you are not looking to solve this by factorisation.

    drpizza hints at this in post #2.
     
    Last edited: Feb 13, 2007
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