SOLUTION: how do you solve? m^2 + m - 72 = 0

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Question 28176: how do you solve? m^2 + m - 72 = 0
Found 2 solutions by Earlsdon, sdmmadam@yahoo.com:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve: Try factoring.
m%5E2+%2B+m+-+72 The factors will have the form: %28m+%2B+a%29%28m+%2B+b%29 where:
a + b = 1 and a*b = -72
Try: a = -8 and b = 9
-8+9 = 1
(-8)(9) = -72
So,
m%5E2+%2B+m+-+72+=+0 Factor.
%28m+-+8%29%28m+%2B+9%29+=+0 Apply the zero products principle.
m-8+=+0 and/or m+%2B+9+=+0
Ifm-8+=+0 then m+=+8
Ifm%2B9+=+0 then m+=+-9
The roots (solutions) are:
x = -8
x = 9

Answer by sdmmadam@yahoo.com(530) About Me  (Show Source):
You can put this solution on YOUR website!
m^2 + m - 72 = 0----(1)
m^2 + [(9m) +(-8m)] - 72 = 0
[m^2+9m]-8m-72=0 (by additive associativity)
m(m+9)-8(m+9) = 0
mp -8p = 0 where p = (m+9)
p(m-8) = 0
(m+9)(m-8)=0
(m+9) = 0 gives m = -9
(m-8) = 0 gives m = 8
Answer: m = -9 and m = 8
Put m=-9 on the LHS of (1)and get 81-9-72 = 0 =RHS
m= 8 on the LHS of (1)and get 64+8-72 = 0 =RHS
Therefore our values m = -9 and m = 8 are correct
NOTE:
[The golden rule is: write the mid term as the sum of two terms whose product is the product of the square term and the constant term]
[HOW? Multiply the square term and the constant term: (m^2)X(-72) = -72m^2
And (-72) = -(1X2X2X2X3X3) = (9)X(-8) so that the mid term: (m) =[(9m) +(-8m)] and the product of these two parts:(9m) and (-8m) = -72m^2 using the concept
If the sign of the product is minus, and the middle term if +,then give the sign of the midterm to the larger set and the other sign to the smaller.