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March 25, 2017

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How do I solve Quadratic equations by factoring?
(for example how would I solve n2 - 10n = -21 if I was supposed to get the answers 7 and 3?

  • Pre algebra - ,

    first, set everything equal to zero.

    n^2 - 10n + 21 = 0

    now, since (n-a)(n-b) = n^2 - (a+b)n + ab, we are looking for two numbers a and b, which multiply to 21 and add to 10.

    Now, the only factor of 21 are 7 and 3. Luckily, they add to 10, so we have

    (n-7)(n-3) = n^2 - 10n + 21

    Now, back to step 1. Why set everything to zero? We now have

    (n-7)(n-3) = 0

    the only way two numbers can multiply to zero is if one or the other of them is zero. So, we have either

    n-7 = 0 ==> n=7
    or
    n-3 = 0 ==> n=3

    Those are the solutions to
    (n-7)(n-3) = 0, which is just a rewriting of the original equation.

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