Question 204320: Three times the square of a certain postive number exceeds six times the number by nine. Find the number?
Found 2 solutions by RAY100, Edwin McCravy:
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! 3n^2 -9 = 6n,,,,,where n = number
.
3n^2 -6n -9 =0
.
(3n+3)(n-3) =0
.
n= -1,,3
.
check
(-1),,3(-1)^2 -9 =6(-1),,,3-9=-6,,,,ok
(3),,,3(3)^2 -9 =6(3),,,,27-9=18,,,,ok
.
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website! Edwin's solution and explanation:
Three times the square of a certain postive number exceeds
six times the number by nine.
Let N be "a certain positive number" as well as "the number"
Three times the square of N exceeds six times N by nine.
Let "the square of N" be N2.
Three times N2 exceeds six times N by nine.
Replace "Three times N2 by 3N2.
3N2 exceeds six times N by nine.
Replace "six times N" by 6N
3N2 exceeds 6N by nine.
That means if you subtract them (larger minus smaller)
you'll get 9.
The larger is the one that does the exceeding, 3N2,
so the smaller one is 6N, so when we subtract them by putting
a minus sign between them, we must get 9.
3N2 - 6N = 9
3N2 - 6N - 9 = 0
3(N2 - 2N - 3) = 0
3(N - 3)(N + 1) = 0
Divide both sides by 3
(N - 3)(N + 1) = 0
N - 3 = 0; N + 1 = 0
N = 3; N = -1
Since the requirement "a certain POSITIVE numbers"
was given we discared the -1 and the only solution
is N = 3
Checking:
Three times the square of 3 exceeds six times 3 by nine.
Three times 9 exceeds six times 3 by nine.
27 exceeds 6 times 3 by 9
27 exceeds 18 by 9
That is true.
Edwin
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