Question 265347: The square of the first of three consecutive odd integers is 9 more than 6 times the sum of the second and the first. Find the numbers.
Found 4 solutions by mananth, ikleyn, josgarithmetic, timofer:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! The square of the first of three consecutive odd integers is 9 more than 6 times the sum of the second and the first. Find the numbers.
let the numbers be x , x+2 , x+4
(x+x+2+x+4)^2 = 6(x+x+2) +9
(3x+6)^2=12x=12+9
9x^2+36x+36=12x+3
9x^2+36x-12x+36-3=0
9x^2+24x+15=0 ( divide equation by3
3x^2+8x+5=0
3x^2+3x+5x+5=0
3x(x+1)+5(x+1)=0
(x+1)(3x+5)=0
x=-1
the numbers are -1 , +1, +3
Answer by ikleyn(53751)  (Show Source):
You can put this solution on YOUR website! .
The square of the first of three consecutive odd integers is 9 more than 6 times the sum of the second and the first.
Find the numbers.
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The solution in the post by @mananth is incorrect. His answer is incorrect, as well.
See below my correct solution.
Let the three consecutive odd integer numbers be x, (x+2) and (x+4).
The equation is
x^2 = 6(x + (x+2)) + 9,
Write it in standard form quadratic equation
x^2 - 12x - 21 = 0. (1)
Its discriminant is
d = b^2 - 4ac = (-12)^2 -4*1*(-21) = 144 + 84 = 228.
The number 228 is not a perfect square.
It tells that equation (1) has no solutions in integer numbers.
The conclusion is that the given problem has no solutions and describes a situation which NEVER may happen.
In other words, the problem as it is given in the post is a FAKE.
Answer by josgarithmetic(39792)  (Show Source):
Answer by timofer(155) (Show Source):
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