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Completing the square problem

  1. Jul 4, 2008 #1
    hi guys.
    If f(x) = x4 + 2x3 + 5x2 - 16x - 20, show that f(x) can be expressed in the form (x2 + x + a)2 - 4(x + b)2, where a and b are constant to be determined.

    Hence, or otherwise, find both the real roots of the eqaution f(x) = 0. Find also the set of values of x such that f(x) > 0.

    I tried completing the square
    but i found
    (x2 + x + (4x+30)/8)2 - 4[(x2)/16 + 4x + 185/32]

    my a and b is WAY too off the answer, which is a = 4 and b = 3. Any help?
     
  2. jcsd
  3. Jul 4, 2008 #2

    Defennder

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    You are given the answer in terms of a,b. So what you have to do is to expand out that answer and simply compare coefficients of powers of x to find out a,b.
     
    Last edited: Jul 4, 2008
  4. Jul 4, 2008 #3
    i have to find a and b first. I think i expand it wrongly. How should i complete the square for x^4 + 2x^3 + 5x^2 - 16x - 20 ? I tried again. I did :

    (x^2 + x + 5/2)^2 - x^2 - 25/4 - 16x - 20. Correct?
     
  5. Jul 4, 2008 #4

    Defennder

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    Did the question tell you to complete the square? If not, you have x^4 + 2x^3 + 5x^2 -16x - 20 = (x^2+x+a)^2 - 4(x+b)^2. All you need to do is to expand out the RHS and compare the coefficients to get the values of a,b.
     
  6. Jul 4, 2008 #5
    Oh ya... that is way easier ... dammit. Thanks.
     
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