Find the solution set Problem in comments
\[\LARGE x^2+14x+49=0\] @mathmale
Aha: a factoring problem. Which of the following methods are you most comfortable with? (a) direct factoring based on the knowledge that 49 factors out to 7*7 or (-7)(-7); (b) quadradic formula, (c) graphing; (d) completing the square? We could go through any of those; your choice.
A seems to be the easiest, or is there an easier method? =]
(a) is the easiest, yes! :) Think: x^2 + 14x + 49 = two binomial (two-term) factors, each with x and each with 7 in it: (x+7)(x+7). You can check the correctness of this result by using the FOIL method to multiply out (x+7)(x+7).
This is a relatively easy problem. If you'd like to discuss a harder one, then by all means post it.
So the solution set is {7,7}?
Never mind, it's -7 :D. Thanks! :)
Actually, you were right the first time: The solution set is indeed {7, 7}. You have TWO solutions which happen to be identical. (That's OK.)
Oh it just told me "-7" was correct and {7,7} was not correct O_o strange?
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