Question 314826: SOLVE BY USING QUADRATIC EQUATION
FIND THE CONSECUTIVE INTEGERS SUCH THAT THE SUM OF THEIR SQUARES IS 245
PLEASE HELP I HAD QUESTIONS LIKE THIS ON MY FINAL BUT I COULD NOT UNDER STAND HOW TO WORK THIS PROBLEM
Found 4 solutions by mananth, MathTherapy, timofer, KMST:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! let the number be x
the consequtive number will be x+1
sum of their squares = x^2 +(x+1)^2= 245
x^2 +(x+1)^2= 245
x^2+x^2+2x+1 =245
2x^2+2x-244=0
x^2+x-122=0
x^2 +12x-11x-122=0
x(x+12)-11(x+12)=0
(x+12(x-11)=0
x= -12 or 11
the numbers are -12 & -11 OR 11 & 12
Answer by MathTherapy(10719)  (Show Source):
You can put this solution on YOUR website!
SOLVE BY USING QUADRATIC EQUATION
FIND THE CONSECUTIVE INTEGERS SUCH THAT THE SUM OF THEIR SQUARES IS 245
PLEASE HELP I HAD QUESTIONS LIKE THIS ON MY FINAL BUT I COULD NOT UNDER STAND HOW TO WORK THIS PROBLEM
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I don't know how the other person can say that these 2 integers are 11 and 12. They are NOT!!
Correct answer: NO 2 integers'/consecutive integers' sum of squares is 245. So, NO SOLUTION!!
Answer by timofer(144)  (Show Source):
You can put this solution on YOUR website! "THE CONSECUTIVE INTEGERS"
How many consecutive integers? The question or description is not complete.
If you want THREE consecutive integers, you should have no trouble finding sum of their squares to be 8, 9, 10.
If middle is n, then the consecutive integers n-1, n, n+1.
Setup and solve  .
Answer by KMST(5336)  (Show Source):
You can put this solution on YOUR website! There are only two sets of consecutive integers with squares that add up to 245.
 and 
The sum of the squares of a set of less than 3 or more than 3 consecutive integers is never 245.
The sum of the squares of 15 or more consecutive integers is always more than 245.
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