Question 1010950: Find two consecutive integers where the sum of the squares of the two integers equals 421.
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! x^2+(x+1)^2= 421
Expand out terms of the left hand side:
2x^2+2x+1 = 421
Divide both sides by 2:
x^2+x+1/2 = 421/2
Subtract 1/2 from both sides:
x^2+x = 210
Add 1/4 to both sides:
x^2+x+1/4 = 841/4
Write the left hand side as a square:
(x+1/2)^2 = 841/4
Take the square root of both sides and, like with every quadratic, we get two possible outcomes:
x+1/2 = 29/2 or x+1/2 = -29/2
Subtract 1/2 from all sides:
x = 14 or x = -15
OK, let's try these and see which one is correct, start with 14:
14^2+(14+1)^2= 421
14^2+15^2= 421
196+225= 421
421= 421 So 14 is our number, and the problem says "consecutive integers", thus the other number is 15
J
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