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Homework Help: Quadratic equation

  1. Mar 1, 2010 #1
    Here is a quadratic equation along with its solution. There's only 1 thing I don't understand. In the brackets how do we know that the values are - 4 and - 5 as opposed to + 4 and + 5. Why are they both negative when the quadratic contains both a negative and a positive expression?

    x2 - 9 + 20 = 0

    (x - 4) (x - 5) = 0

    x = 4 or 5
     
  2. jcsd
  3. Mar 1, 2010 #2
    Let me ask you a question.
    If we have ab=0 ,
    what does it suggest?
    [at least either a or b has to be 0]
     
  4. Mar 1, 2010 #3
    What happens when you multiply two negatives vs. adding two negatives?
     
  5. Mar 1, 2010 #4

    statdad

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    Homework Helper

    I'll assume your original quadratic should read [tex] x^2 - 9x + 20 = 0 [/tex].

    When you have a quadratic with leading coefficient equal to 1, and integer coefficients, there is a simple process to follow to check whether the polynomial will or will not factor with integer constants.
    Step 1: Write down all pairs of integers with product equal to the constant term.

    Step 2: Look in your list for a pair whose SUM equals the middle coefficient. These are the appropriate choices. (IF there is no such pair, the given quadratic won't factor this way)

    For yours, the constant is 20. There are two possible choices: [tex] 4 \text{ and } 5 [/tex] and [tex] -4 \text{ and } -5 [/tex]. Since the second pair sum to [tex] -9 [/tex], these are the choices, and you know

    [tex]
    x^2 - 9x + 20 = (x-4)(x-5)
    [/tex]

    As another example, consider factoring

    [tex]
    x^2 - 6x - 27 = 0
    [/tex]

    Step 1: Look for integer pairs that multipy to [tex] -27[/tex]. There are two choices: [tex] 3 \text{ and } -9 [/tex], and [tex] -3 \text{ and } 9 [/tex]. The first pair has sum [tex] -6 [/tex], so those are your choices.

    [tex]
    x^2 - 6x- 27 = 0 \Rightarrow (x-9)(x+3) = 0
    [/tex]

    so the solutions are [tex] x = 9 \text{ and } x = -3 [/tex].

    Finally, consider [tex] x^2 - 11x + 8 = 0 [/tex]. Sets of integers with a product of [tex] 8 [/tex] are [tex] 1 \text{ and } 8, 2 \text{ and } 4, -1 \text{ and } -8, -2 \text{ and } -4[/tex]. None of these pairs sum to [tex] -11 [/tex], so this polynomial doesn't factor using integers.

    It is very important to remember that this method works ONLY when the leading coefficient (the coefficient of [tex] x^2 [/tex]) equals 1.
     
  6. Mar 2, 2010 #5

    HallsofIvy

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    Science Advisor

    [itex](x- a)(x- b)= x^2+ (-a-b)x+ (-a)(-b)

    (-a-b)= -(a+b) but (-a)(-b)= ab- the sum of two negative numbers is negative but the product of two negative numbers is positive.

    -4-5= -9, (-4)(-5)= +20.
     
  7. Mar 5, 2010 #6
    Thanks a lot everyone. It's a lot clearer now.
     
  8. Mar 5, 2010 #7
    Thanks to help from various people on this site I have learned how to solve trinomial quadratics when the coefficient of x2 is 1. However, I've now moved on to this type of quadratic. It seems that a different approach is needed:

    2x2 - 7x - 15 = 0

    I know from looking at the answers, that the factors required are as follows:
    (2x + 3) (x - 5) = 0
    But how did we arrive at + 3 and - 5? They don't add to make -7.
     
  9. Mar 5, 2010 #8

    Dembadon

    User Avatar
    Gold Member

    For quadratic equations where the coefficient of x2 is not 1, I like to use the "AC method."

    ax2 + bx + c = 0

    In the above equation, a, b, and c represent constants and x is the variable. First, multiply the a term and the c term. Then find two numbers whose sum is b and whose product is ac. After this you can factor by grouping after rewriting the equation using the two numbers you obtained in the previous step.

    m + n = b
    (m)(n) = (a)(c)

    (ax2 + mx) + (nx + c) = 0
     
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