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Math Help - Simple Algebra Problem

  1. #1
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    Simple Algebra Problem

    I am having trouble with simple algebra. It has been 27 years since I last encountered any real math, and I am in the process of reviewing to help my grandson. Can I please get step-by step instructions on how to solve this simple math problem?:

    x^2+5x+9=5

    Any help is appreciated.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Emodius View Post
    I am having trouble with simple algebra. It has been 27 years since I last encountered any real math, and I am in the process of reviewing to help my grandson. Can I please get step-by step instructions on how to solve this simple math problem?:

    x^2+5x+9=5

    Any help is appreciated.
    an equation of this type is called a "quadratic equation." there are three main ways to solve it.

    1) By the quadratic formula ---will work in any case
    2) By completing the square ---will work in any case
    3) By foiling ---works with "nice" quadratics, but is generally simpler so use when you can

    which method is your grandson currently learning?
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Emodius View Post
    I am having trouble with simple algebra. It has been 27 years since I last encountered any real math, and I am in the process of reviewing to help my grandson. Can I please get step-by step instructions on how to solve this simple math problem?:

    x^2+5x+9=5

    Any help is appreciated.
    Three methods:
    1. Factoring
    x^2 + 5x + 9 = 5

    x^2 + 5x + 4 = 0

    Multiply the coefficient of the x^2 term (1) and the constant term (4): 1 x 4 = 4

    Now write out all the factors of 4:
    1, 4 <-- This also covers 4, 1
    2, 2
    -1, -4
    -2, -2

    Now look for the pair that adds to the coefficient of the linear term (5):
    1 + 4 = 5
    2 + 2 = 4
    -1 + -4 = -5
    -2 + -2 = -4

    So we need the 1, 4 pair.

    Now write the equation as the following:
    x^2 + 5x + 4 = 0

    x^2 + (1x + 4x) + 4 = 0

    (x^2 + x) + (4x + 4) = 0

    Now factor the common factor from each set of parenthesis:
    x(x + 1) + 4(x + 1) = 0

    Now each term has a common factor, the x + 1. So factor this from the overall expression:
    (x + 4)(x + 1) = 0

    Now, if this is to be true, one of (or both of) the factors must be 0.

    Thus x + 4 = 0 ==> x = -4
    or
    x + 1 = 0 ==> x = -1

    Thus the solution is x = -4, -1.

    (This method works for any problem that factors. If it doesn't factor then you won't be able to find a pair of numbers that add to be the coefficient of the linear term. In that case you need to use one of the other two techniques.)

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Emodius View Post
    I am having trouble with simple algebra. It has been 27 years since I last encountered any real math, and I am in the process of reviewing to help my grandson. Can I please get step-by step instructions on how to solve this simple math problem?:

    x^2+5x+9=5

    Any help is appreciated.
    2. Completing the square.

    x^2 + 5x + 9 = 5

    x^2 + 5x = -4

    Compare the LHS of the equation with the LHS of the following:
    x^2 + 2ax + a^2 = (x + a)^2

    We see that if we find the right a value then all we need to do is add a^2 to both sides to make the LHS a perfect square.

    So compare the two linear coefficients:
    5 = 2a ==> a = 5/2

    Thus we need to add a^2 = 25/4 to both sides of the equation:
    x^2 + 5x = -4

    x^2 + 5x + 25/4 = -4 + 25/4

    (x + 5/2)^2 = -4 + 25/4 = -16/4 + 25/4 = 9/4

    (x + 5/2)^2 = 9/4

    x + 5/2 = (+/-)sqrt{9/4} = (+/-)(3/2)

    x = -5/2 (+/-) 3/2

    So
    x = -5/2 + 3/2 = -2/2 = -1
    or
    x = -5/2 - 3/2 = -8/2 = -4

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Emodius View Post
    I am having trouble with simple algebra. It has been 27 years since I last encountered any real math, and I am in the process of reviewing to help my grandson. Can I please get step-by step instructions on how to solve this simple math problem?:

    x^2+5x+9=5

    Any help is appreciated.
    3. The quadratic formula
    x^2 + 5x + 4 = 0

    If we have the trinomial ax^2 + bx + c = 0 then
    x = [-b (+/-) sqrt{b^2 - 4ac}]/[2a]

    In this case a = 1, b = 5, c = 4.

    x = [-5 (+/-) sqrt{5^2 - 4*1*4}]/[2*1]

    x = [-5 (+/-) sqrt{25 - 16}]/2

    x = [-5 (+/-) sqrt{9}]/2

    x = [-5 (+/-) 3]/2

    So
    x = [-5 + 3]/2 = [-2]/2 = -1
    or
    x = [-5 - 3]/2 = [-8]/2 = -4
    as you might have expected by now.

    -Dan
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  6. #6
    Newbie Logarithm Freak's Avatar
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    Use this

    x= (-b +/- (b^2 -4ac)^1/2)/(2a)
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Logarithm Freak View Post
    Use this

    x= (-b +/- (b^2 -4ac)^1/2)/(2a)
    Don't just give an answer like this explain where it comes from, and for that
    matter what the variables/symbols represent.

    RonL
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