SOLUTION: The sum of the squares of two consecutive integers is thirty one less than twice the square of the second integer. What are the integers? I tried this one: (x+1) (x+2) and then

Question 761695: The sum of the squares of two consecutive integers is thirty one less than twice the square of the second integer. What are the integers?
I tried this one: (x+1) (x+2) and then, (x+x+1+2)= 2x+3....
and I haven't gone farther because I was looking if i have to take thirty one or square it first?
THIS IS URGENT!
Thanks,
Louise
Found 2 solutions by ramkikk66, stanbon:
Answer by ramkikk66(644) (Show Source): You can put this solution on YOUR website!
The sum of the squares of two consecutive integers is thirty one less than twice the square of the second integer. What are the integers?

I think you got stuck because you have not actually *squared* the numbers in the expressions you wrote.
Solution:
Let the 2 integers be x and x+1.
Square of the 1st integer =
Square of the 2nd integer =
Sum of the squares = (expression 1)
Twice the square of second integer = (expression 2)
Sum of the squares (expression 1) is 31 less than twice the square of the second number (expression 2)
So
Simplifying and cancelling out the 2*x^2 on both sides

or

The two numbers are 15 and 16.

Sum of squares = 225 + 256 = 481.
Twice square of 16 = 512.
512 - 481 = 31. Check!
:)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The sum of the squares of two consecutive integers is thirty one less than twice the square of the second integer. What are the integers?
I tried this one: (x+1) (x+2) and then, (x+x+1+2)= 2x+3....
and I haven't gone farther because I was looking if i have to take thirty one or square it first?
-------------------------
1st: x
2nd: x+1
----------
x^2+(x+1)^2 = 2(x+1)^2-31
x^2 + x^2 + 2x + 1 = 2[x^2+2x+1] - 31
-----
2x^2 + 2x + 1 = 2x^2 + 4x + 2 -31
------
2x+1 = 4x - 29
2x = 30
x = 15
---------------------
x+1 = 16
=======================
Cheers,
Stan H.

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