SOLUTION: Please help me set up. A sguare of two consecutive integers exceeds the sum of their sguares by 84.

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Question 822477: Please help me set up. A sguare of two consecutive integers exceeds the sum of their sguares by 84.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39623)   (Show Source): You can put this solution on YOUR website!
Consecutive integers, I and I+1.

Why do you say, "A square of..."? This and the rest of your description are confusing. Write the exact problem, word for word.

I will try with what you have presented in the current request.
(I+1)^2, a square of one of the integers. This would be the larger one.
Sum of the squares of the integers, (I)^2+(I+1)^2.
.
Exceeds.... by 84?
(I+1)^2-I^2-(I+1)^2=84, meaning their difference is 84, expecting I+1 to be the larger integer.
I^2+2I+1-I^2-(I^2+2I+1)=84
-I^2=84
NO GOOD!

Give the exact problem statement exactly as was given to you.
Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!

Please help me set up. A sguare of two consecutive integers exceeds the sum of their sguares by 84. 


Let first integer be F

Then second = F + 1

Therefore, 

Solve to determine the value of F, and then F + 1



Then do a check!! 



Send comments, “thank-yous,” and inquiries to “D” at MathMadEzy@aol.com.

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