Question 1197907: Find two consecutive negative odd integers such that 5 times the first plus the square of the second equals -14. Found 4 solutions by josgarithmetic, ewatrrr, MathTherapy, greenestamps:Answer by josgarithmetic(39613) (Show Source):
You can put this solution on YOUR website! two consecutive negative odd integers: n and (n+2)
Question states: Write as You Read.
5n + (n+2)^2 = -14
5n + n^2 + 4n + 4 = -14
n^2 + 9n + 18 = 0
(n +3)(n+6) = 0
n = -3 0r n = -6
{-3, -1 } 0r { -6, - 4}
You can put this solution on YOUR website!
Find two consecutive negative odd integers such that 5 times the first plus the square of the second equals -14.
(1) Never solve a quadratic equation the way tutor @josgarthmetic did:
looks like...
(2) The solution from tutor @ewatrrr is algebraically sound; but the problem specified that the two integers are odd.
To avoid overlooking the fact that the problem asks for consecutive odd integers, you can use "2n-1" and "2n+1" for the two integers; then an integer solution for n will guarantee that the two integers are odd.
or
n must be an integer, so we reject the solution n = -5/2; the single solution to our equation is n = -1.
That gives us, for the two consecutive negative odd integers that we are looking for,
2n-1 = 2(-1)-1 = -2-1 = -3
2n+1 = 2(-1)+1 = -2+1 = -1
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