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Math Help - a^2 + 2ab + b^2

  1. #1
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    Smile a^2 + 2ab + b^2

    Hey, think anyone could kind of help me understand a^2 + 2ab + b^2 better?
    I simply just can not figure out alot of the larger problems such as:

    4k^2 -36k + 81

    I know the answer is (2k - 9)^2 but dont understand how to get that using the formula above. Well except its - instead of the + from above. If anyone could possibly work a few random examples or even the one above it might help me understand the concept of the a^2 + 2ab + b^2 which I cant seem to figure out at all. ^_^

    Thanks
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  2. #2
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    Hello

    Quote Originally Posted by Shrinkwrap View Post
    Hey, think anyone could kind of help me understand a^2 + 2ab + b^2 better?
    I simply just can not figure out alot of the larger problems such as:

    4k^2 -36k + 81

    I know the answer is (2k - 9)^2 but dont understand how to get that using the formula above. Well except its - instead of the + from above. If anyone could possibly work a few random examples or even the one above it might help me understand the concept of the a^2 + 2ab + b^2 which I cant seem to figure out at all. ^_^

    Thanks
    The formula is

    (a-b)^2 = a^2-2ab+b^2

    and it is 4k^2 -36k + 81

    compare these to terms

    4k^2 = a^2
    -2ab = -36k
    b^2 = 81


    4k^2 = a^2 => a = 2k
    81 = b^2 => b=9

    Now use the second equality: -2ab = -36k to verify your solution

    -2*a*b = -2*2k *9 = -4k*9 = -36k

    => (a+b)^2 = (2k-9)^2
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  3. #3
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    i kind of get it... im still confused though, such as if a could not be perfectly squared. a few of my problems require me to complete the square but using the a^2 + 2ab + b^2 rule I come out with square root numbers which cant be right:

    examples of two are:
    5k^2 + k - 2 = 0

    I come out with 20k^2 + 4k + 4 = 12 and while ignoring the right side, I cant come up with any correct perfect square using the formula.

    3a^2 - 1 = 6a

    Same thing with this one, I come out with 3a^2 + 6a + 9 = 10 which once you try to do the 3a^2 = a^2, I dont know how u get a non rooted number.
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  4. #4
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    Quote Originally Posted by Shrinkwrap View Post
    i kind of get it... im still confused though, such as if a could not be perfectly squared. a few of my problems require me to complete the square but using the a^2 + 2ab + b^2 rule I come out with square root numbers which cant be right:

    examples of two are:
    5k^2 + k - 2 = 0

    I come out with 20k^2 + 4k + 4 = 12 and while ignoring the right side, I cant come up with any correct perfect square using the formula.
    5k^2 +k - 2 = 0 (1)
    First devide by 5

    k^2 + \frac{1}{5}k - \frac{2}{5}

    completing the square: add 0 = (\frac{1}{5*2})^2 - (\frac{1}{5*2})^2

    Thus

    k^2 + \frac{1}{5}k - \frac{2}{5} +(\frac{1}{5*2})^2 - (\frac{1}{5*2})^2

    k^2 + \frac{1}{5}k - \frac{2}{5} +(\frac{1}{10})^2 - (\frac{1}{10})^2

    k^2 + \frac{1}{5}k - \frac{2}{5} +(\frac{1}{10})^2 - (\frac{1}{100})

    k^2 + \frac{1}{5}k +(\frac{1}{10})^2 - (\frac{1}{100}) - \frac{2}{5} = 0

    k^2 + \frac{1}{5}k +(\frac{1}{10})^2  = (\frac{1}{100}) + \frac{2}{5}

    (k+\frac{1}{10})^2 = (\frac{1}{100}) + \frac{2}{5}

    First you consider ax^2+bx+c

    Devide by a => x ^2+\frac{bx}{a}+\frac{c}{a}

    Define \frac{b}{a} =: B, means

    x^2+Bx+\frac{c}{a}

    then add +(\frac{B}{2})-(\frac{B}{2})^2

    This is equal to this: +(\frac{B}{2})-(\frac{B}{2})^2 <br />
  = (\frac{b}{a*2})^2 -(\frac{b}{a*2})^2

    But i think its hard to read

    I guess, this is not correct.
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  5. #5
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    to make the square u first need to get rid of the co effiecent of the x^2 term and isolate the x^2 and x terms


    5k^2 +k= 2 (add 2 to each side)
    5(k^2 +1/5k)=2
    k^2+k = 2/5 (divide by 5 on both sides)
    k^2+k+1/4= 1/4+2/5 = 5/20+8/20 =13/20
    (k+1/2)^2 = 13/20
    k+1/2 = sqr13/20
    k=-1/2+-sqr13/20
    k=-1/2-sqr13/20

    ok i dont know if this is 100% correct probably 100% wrong but to make the perfect square u need to add half the co effiecent of x then square that number

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


    3a^2 -6a-1= 0
    3a^2-6a= 1
    3(a^2-2a) = 1
    a^2-2a = 1/3
    a^2-2a+(1)^2 = 1/3 (half a 2a squared)
    a^2-2a+1 = 1/3
    (a-1)^2 = 1/3
    a-1=sqr1/3 (get the square root of both sides)
    a=1+sqr1/3
    a=1-sqr1/3

    i may not understand what you are trying to do so u try to solve one so i can see how you been working it out and see if my way is correct or not?
    i am teaching myself how to do these so my answers might be wrong but im sure my working out is right for completing the perfect square and im sure the answer if right can be simplified
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