Intermedia Algebra
posted by Elisa
Solve by factoring:
2a^2 − 3a = −5

Henry
2a^2  3a = 5.
2a^2  3a  5 = 0,
Use the AC Method:
A*C = 2*(5) = 10,
2a^2 + (2a  5a) 5 = 0,
Group the 4 terms into 2 factorable pairs:
(2a^2 + 2a) + (5a  5) = 0,
Factor each binomial:
2a(a + 1) 5(a + 1) = 0,
Factor out a+1:
(a + 1)(2a  5) = 0,
a + 1 = 0,
a = 1.
2a  5 = 0,
2a = 5,
a = 2.5.
Solution set: a = 1, and a = 2.5.
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