SOLUTION: Find three consecutive integers such that the square of the sum of the smaller two is 105 more than the square of the largest.

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Question 1162614: Find three consecutive integers such that the square of the sum of the smaller two is 105 more than the square of the largest.
Found 3 solutions by josgarithmetic, ikleyn, MathTherapy:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Integers could be as n-1, n, n+1.

%28%28n-1%29%2Bn%29%5E2=105%2B%28n%2B1%29%5E2
-
%282n-1%29%5E2=105%2Bn%5E2%2B2n%2B1
4n%5E2-4n%2B1=105%2Bn%5E2%2B2n%2B1
3n%5E2-4n-2n=105
3n%5E2-6n=105
n%5E2-2n=35
n%28n-2%29=35
7%2A5=35
Middle Number could be 7.
Largest Number, 8.

Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
.

Integers could be as  n-1, n, n+1.


%28%28n-1%29%2Bn%29%5E2=105%2B%28n%2B1%29%5E2
-
%282n-1%29%5E2=105%2Bn%5E2%2B2n%2B1

4n%5E2-4n%2B1=105%2Bn%5E2%2B2n%2B1

3n%5E2-4n-2n=105

3n%5E2-6n=105

n%5E2-2n=35

n%28n-2%29=35


This equation has TWO SOLUTIONS:  n = 7  and  n = -5.


So, there are TWO OPPORTUNITIES for the triple of the numbers:


    one triple is         6,  7,  8.

    the other triple is  -6, -5, -4.


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Find three consecutive integers such that the square of the sum of the smaller two is 105 more than the square of the largest.
Let first number be N

Then 2nd and 3rd are: N + 1, and N + 2, respectively

We then get: 

First number or highlight_green%28matrix%281%2C3%2C+N%2C+%22=%22%2C+%22%22%2B-+6%29%29

Numbers are: