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December 22, 2014

December 22, 2014

Posted by **Grace** on Thursday, June 7, 2012 at 9:21pm.

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

- Pre algebra -
**Steve**, Friday, June 8, 2012 at 11:31amfirst, 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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