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Solve 2x^2+12x+18=0?

by Guest58202  |  earlier

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Solve 2x^2+12x+18=0?

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  1. factor out 2   -->  2( x^2 +6x +9 ) = 0

    factor the polynomial    (x+3) (x+3) = X^2 +6x +9)

    therefore 2 (x+3) (X+3) = 0     x =  - 3


  2. (2x+6)(x+3)=0

    x=-3

    make it a good day

  3. 2x^2+12x+18=0

    4x+12x+18=0

    16x+18=0

    16x=-18

    x=-1.125

  4. 2x^2 + 12x + 18 = 0

    (2x^2 + 12x + 18)/2 = 0/2

    x^2 + 6x + 9 = 0

    x^2 + 3x + 3x + 9 = 0

    (x^2 + 3x) + (3x + 9) = 0

    x(x + 3) + 3(x + 3) = 0

    (x + 3)(x + 3) = 0

    x + 3 = 0

    x = -3

    ∴ x = -3

  5. =(2X+6) (X+3)

    = X= -3

  6. x² + 6x + 9 = 0

    (x + 3)(x + 3) = 0

    x = - 3

  7. 2x^2 + 12x +18 = 0

    (2x + 6) (1x + 3) = 0

    (2x +6) = 0 and (1x +3) = 0

    2x = -6 and x = -3

    x = -3


  8. Either use quadratic formula or factor.

    I'll factor...

    Divide everything by 2 first to make it easier...

    2 ( x^2 + 6x + 9) = 0

    2 ( x + 3) (x + 3 ) = 0

    Use zero product property

    Solution is x = - 3.

    The 2 on the outside doesn't matter because 2 does not equal zero.

    The solution of x = -3 is called a double root because the solution of -3 occurred twice.

    Good luck to you !
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