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Math Help - Quadratic Equation general question

  1. #1
    Member cmf0106's Avatar
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    Quadratic Equation general question

    As we know the standard form of a quadratic equation is ax^2 + bx + c = 0

    But when B is equal to 0 in the above it looks like this ax^2 =c. However to get the c in the preceding to look like a standard quadratic equation you must add its inverse to the other side.

    Take this for example

    x^2 = 49 Obviously you could just do the \sqrt49 to get +- 7. But What if you wanted to put it in general quadratic form?

    If I wanted to put it general quadratic form would this be correct?
    x^2 - 49 = 0
    (x + 7)(x - 7) = 0

    Then just use the multiplicative property of zero to solve both equations and arrive at +- 7?

    So in conclusion has my reasoning and calculations been correct up to this point?

    Many thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cmf0106 View Post
    As we know the standard form of a quadratic equation is ax^2 + bx + c = 0

    But when B is equal to 0 in the above it looks like this ax^2 =c. However to get the c in the preceding to look like a standard quadratic equation you must add its inverse to the other side.

    Take this for example

    x^2 = 49 Obviously you could just do the \sqrt49 to get +- 7. But What if you wanted to put it in general quadratic form?

    If I wanted to put it general quadratic form would this be correct?
    x^2 - 49 = 0
    (x + 7)(x - 7) = 0

    Then just use the multiplicative property of zero to solve both equations and arrive at +- 7?

    So in conclusion has my reasoning and calculations been correct up to this point?

    Many thanks!
    Um, yeah, pretty much
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  3. #3
    MHF Contributor Mathstud28's Avatar
    Joined
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    Pennsylvania
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    Quote Originally Posted by cmf0106 View Post
    As we know the standard form of a quadratic equation is ax^2 + bx + c = 0

    But when B is equal to 0 in the above it looks like this ax^2 =c. However to get the c in the preceding to look like a standard quadratic equation you must add its inverse to the other side.

    Take this for example

    x^2 = 49 Obviously you could just do the \sqrt49 to get +- 7. But What if you wanted to put it in general quadratic form?

    If I wanted to put it general quadratic form would this be correct?
    x^2 - 49 = 0
    (x + 7)(x - 7) = 0

    Then just use the multiplicative property of zero to solve both equations and arrive at +- 7?

    So in conclusion has my reasoning and calculations been correct up to this point?

    Many thanks!
    Yes, up to now your conclusion has been correct, but even more generally

    in ax^2+b+c=0 if we let b=0

    then by the quadratic formula the solution is

    \frac{0\pm\sqrt{0-4(a)(c)}}{2a}

    Which shows the fact that for a case where b=0 one of the two must have a negative coefficient otherwise ti would yield an imaginary result
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