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Math Help - solving quadratic equation by completing the square

  1. #1
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    Exclamation solving quadratic equation by completing the square

    HELP NEEDED

    solve the following by completing the square

    2x sqaured (dont know to insert power of 2) + 4x - 6

    i know i have to divide it by 2 so i get

    x sqaured + 2x - 3

    but i dont knwo how to work it

    also once i have got the answer i have to check it how do i you do this


    PLEASE I REALLY NEED HELP!!
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  2. #2
    MHF Contributor
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    Hi

    2x^2 + 4x - 6 = 2(x^2 + 2x - 3) = 2(x^2 + 2x + 1 - 4) = 2 \left[(x+1)^2 - 4)\right]
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  3. #3
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    Quote Originally Posted by jm16 View Post
    HELP NEEDED

    solve the following by completing the square

    2x sqaured (dont know to insert power of 2) + 4x - 6

    i know i have to divide it by 2 so i get

    x sqaured + 2x - 3

    but i dont knwo how to work it

    also once i have got the answer i have to check it how do i you do this


    PLEASE I REALLY NEED HELP!!
    I don't understand what you are asked to solve(?). I guess that you should factor the given term:

    2x^2 + 4x - 6 = 2(\color{red}x^2+2x\color{black}-3)

    Use the red term to complete the square:

    2(x^2+2x \bold{{\color{blue}+ 1 - 1}}-3) = 2(\underbrace{(x+1)^2-4}_{\text{difference of squares}})

    A difference of squares can be factored immediately:

    2((x+1)^2-4) = 2((x+1+2)(x+1-2)) = 2(x+3)(x-1)
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jm16 View Post
    HELP NEEDED

    solve the following by completing the square

    2x sqaured (dont know to insert power of 2) + 4x - 6

    i know i have to divide it by 2 so i get

    x sqaured + 2x - 3

    but i dont knwo how to work it

    also once i have got the answer i have to check it how do i you do this


    PLEASE I REALLY NEED HELP!!
    unless there's an "= 0" in your problem, you cannot just divide by 2 like that, youwould have to factor out a 2 in that case. the thing to note about complete squares is that the constant term is always \left( \frac {\text{the coefficient of }x}2 \right)^2. so we normally add and subtract this term (so that we effectively add 0 and hence do not change the problem) and then simplify to get the complete sqaure.

    i will assume we have 2x^2 + 4x - 6 = 0

    \Rightarrow x^2 + 2x - 3 = 0

    \Rightarrow x^2 + 2x {\color{red} + 1^2 - 1^2} - 3 = 0

    (Note that x^2 + 2x + 1 is a complete square)

    \Rightarrow (x + 1)^2 - 1 - 3 = 0

    \Rightarrow (x + 1)^2 - 4 = 0

    can you finish?

    EDIT: Pfft! too late!
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