SOLUTION: the sum of the squares of two consecutive even integers 1252. find the inetegers

Question 45041: the sum of the squares of two consecutive even integers 1252. find the inetegers
Found 2 solutions by Nate, abhijitvakil:
Answer by Nate(3500) (Show Source): You can put this solution on YOUR website!
First Integer: i
Second Integer: i + 2

or
Second Integer: i + 2 = -26 + 2 = -24
Second Integer: i + 2 = 24 + 2 = 26
Answer: -26 and -24
of
Answer: 24 and 26
Answer by abhijitvakil(7) (Show Source): You can put this solution on YOUR website!
Let the lesser integer be x. Then the other number being consecutive even integer to x, becomes (x + 2). Now the relation given in the problem can be stated as :
x^2 + (x + 2)^2 = 1252
or
x^2 + (x^2 + 2*x*2 + 2^) = 1252 ( we use the formula (a + b)^2 = a^2 + 2ab + b^2)
or
x^2 + x^2 + 4x + 4 = 1252
or
2x^2 + 4x + 4 - 1252 = 0
or
x^2 + 2x - 624 = 0 ..... (1) ( we divide the entire equation by 2)
The above is a typical quadratic equation of the type ax^2 + bx + c = 0. Thus we have
a = 1, b = 2 & c = 624
The solution to this equation is given by the following :

substituting the values of a,b, & c into the above, we get two "roots" or solutions to x ( i am skipping the numerical calculations, which the student can do himself ):
x = (24, -26)
Thus if x = 24, then the other integer will be x+2 i.e. 26, and if x = -26, then the other integer will be -26+2 i.e. -24.

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