The sum of the squares of 2 consecutive negative integers is 41. What are the numbers?
Which of the following equations is the result of using the factoring method to solve the problem?
(n - 5)(n - 4) = 0
(n - 5)(n + 4) = 0
(n + 5)(n - 4) = 0
(n + 5)(n + 4) = 0
There are two factors of -36 such that one factor is 11 less than half of the other factor. Choose all the pairs of these factors.
-2 and 18
-6 and 6
3 and -12
4 and -9
+130085 First one....we have that
n^2 + ( n + 1)^2 = 41
n^2 + n^2 + 2n + 1 = 41
2n^2 + 2n - 40 = 0
n^2 + n - 20 = 0
(n + 5) ( n - 4) = 0
Second one
Call the first factor, F
So...the second is F/2 - 11
So.....this implies that
(F) [ (F/2) - 11] = -36 simplify
F^2/2 - 11F = - 36
F^2/2 - 11F + 36 = 0 multiply through by 2
F^2 - 22F + 72 = 0
(F - 18) ( F - 4) = 0
F = 18 or F = 4
When F = 18 the other factor is : F/2 - 11 = 18/2 - 11 = 9 - 11 = -2
When F = 4 the other factor is 4/2 - 11 = 2 - 11 = -9
So....the factors are
(18, -2) and ( 4, - 9)
Which of the following equations is the result of using the factoring method to solve the problem?
(n - 5)(n - 4) = 0
(n - 5)(n + 4) = 0
(n + 5)(n - 4) = 0
(n + 5)(n + 4) = 0
(n - 5)(n - 4) = 0, This comes from this quadratic: n^2 - 9 n + 20 = 0
(n - 5)(n + 4) = 0, This comes from this quadratic: n^2 - n - 20 = 0
(n + 5)(n - 4) = 0, This comes from this quadratic: n^2 + n - 20 = 0
(n + 5)(n + 4) =0, This comes from this quadratic: n^2 + 9 n + 20 = 0
