Asked by Grace
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?
(for example how would I solve n2 - 10n = -21 if I was supposed to get the answers 7 and 3?
Answers
Answered by
Steve
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.
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.