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Math Help - complete the square

  1. #1
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    complete the square

     <br />
y=3x^2+12x-7<br />
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  2. #2
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    Re :

    Quote Originally Posted by william View Post
     <br />
y=3x^2+12x-7<br />

    3x^2+12x-7=0
    x^2+4x-7/3=0
    (x+2)^2=7/3+(2)^2
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    3x^2+12x-7=0
    x^2+4x-7/3=0
    (x+2)^2=7/3+(2)^2
    I don't understand what you did. What is the vertex?
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  4. #4
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    Quote Originally Posted by william View Post
    I don't understand what you did. What is the vertex?
    y = 3x^2 + 12x - 7

    Take out the common factor of 3

    y = 3[x^2 + 4x - \frac{7}{3}]

    Add a cleverly disguised 0... in other words, \left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2

    y = 3[x^2 + 4x + 2^2 - 2^2 - \frac{7}{3}]

    Can you see that x^2 + 4x + 2^2 is a perfect square?

    y = 3[(x + 2)^2 - \frac{19}{3}]

    Expand the common factor of 3 through the brackets.

    y = 3(x+2)^2 - 19


    This is now of the form y = a(x-h)^2 + k where h and k are the x and y co-ordinates of the vertex.

    So what is the vertex?
    Last edited by Prove It; December 2nd 2008 at 12:53 AM.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    y = 3x^2 + 12x - 7

    Take out the common factor of 3

    y = 3[x^2 + 4x - \frac{7}{3}]

    Add a cleverly disguised 0... in other words, \left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)

    y = 3[x^2 + 4x + 2^2 - 2^2 - \frac{7}{3}]

    Can you see that x^2 + 4x + 2^2 is a perfect square?

    y = 3[(x + 2)^2 - \frac{19}{3}]

    Expand the common factor of 3 through the brackets.

    y = 3(x+2)^2 - 19


    This is now of the form y = a(x-h)^2 + k where h and k are the x and y co-ordinates of the vertex.

    So what is the vertex?
    V: (-2,-19)
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