SOLUTION: FIND TWO CONSECUTIVE INTEGERS IF TWO TIMES THE SQUARE OF THE FIRST IS EQUAL TO 23 MORE THAN THE SQUARE OF THE SECOND... thank you.

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: FIND TWO CONSECUTIVE INTEGERS IF TWO TIMES THE SQUARE OF THE FIRST IS EQUAL TO 23 MORE THAN THE SQUARE OF THE SECOND... thank you.      Log On


   

Question 182485: FIND TWO CONSECUTIVE INTEGERS IF TWO TIMES THE SQUARE OF THE FIRST IS EQUAL TO 23 MORE THAN THE SQUARE OF THE SECOND... thank you.
Found 2 solutions by tvandenberg, solver91311:
Answer by tvandenberg(45) About Me  (Show Source):
You can put this solution on YOUR website!
The are n and m=n%2B1.
We know, from the problem:
2%2An%5E2=m%5E2%2B23
Subbing in the information from above,
2%2An%5E2=%28n%2B1%29%5E2%2B23
Some algebra,
2%2An%5E2=n%5E2%2B2n%2B1%2B23
--->
n%5E2-2%2An-24=0
--->
%28n-6%29%28n%2B4%29=0
--->
n=6%2C-4
I would assume it has to be positive, n = 6, m = 7
Checking, 2%2A6%5E2+=+72 7%5E2%2B23=72
Hope this helps.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!

PLEASE DO NOT TYPE YOUR QUESTION IN ALL CAPITALS. IT IS THE ELECTRONIC EQUIVALENT OF SHOUTING, THEREFORE BOTH RUDE AND ANNOYING.


If the first integer is x, then the next consecutive integer is x + 1.

Two times the square of the first integer:

23 more than the square of the second integer:

These two expressions are equal so:







and

, so:



Hence:

or



Check:



Checks.



Checks.

So the required integers are either -4 and -3 or 6 and 7.

John