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Algebra II

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For the : ax^2 + bx + c
How would you get what to put in
(_+_)(_+_)= 0

Is it c doubled and then finding the factors that could equal b or 1/2 of b times c...?

Could someone post how I could find the answer this way (shown above)
(AND NOT THE QUADRATIC FORMULA)?

Thnak You.

  • Algebra II -

    ax^2 + bx + c
    is NOT an equation.

    Why should you expect to be able to rewrite it as (_+_)(_+_)= 0 ?

    The values of x that satisfy the equation
    ax^2 + bx + c = 0
    ARE given by the quadratic formula.

    Call them x1 and x2.

    (x-x1)(x-x2) = 0

    x1 = (1/2a)[-b + sqrt(b^2-4ac)]
    x2 = (1/2a)[-b - sqrt(b^2-4ac)]

  • Algebra II -

    a ( x^2 + (b/a) x + (c/a))

    a ( x^2 + 2(b/(2a)) x + (c/a))

    a [ x^2 + 2(b/(2a)) x + (b/(2a))^2 +
    - ( (b/(2a))^2 - (c/a) )]


    a [( x + (b/(2a)))^2 -
    - ( (b/(2a))^2 - (c/a) )]

    a [( x + (b/(2a)))^2 -
    - ( (b^2 - 4ac/((2a)^2)]


    a [( x + (b/(2a)))^2 -
    - ( (b^2 - 4ac/((2a)^2)]


    a [( x + (b/(2a))) - ( sqrt(b^2 - 4ac/((2a))] *
    [( x + (b/(2a))) + ( sqrt(b^2 - 4ac/((2a))]

    a [( x + ((b/(2a)) - sqrt(b^2 - 4ac/((2a))] *
    [( x + (b/(2a))) + sqrt(b^2 - 4ac/((2a))]

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