SOLUTION: find two consecutive whole numbers such that the sum of their square is 41.

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Question 515261: find two consecutive whole numbers such that the sum of their square is 41.
Answer by Maths68(1474) About Me  (Show Source):
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Let
One number = x
Other number = x+1
sum of their square = 41
x^2+(x+1)^=41
x^2+x^2+2x+1=41
2x^2+2x+1-41=0
2x^2+2x-40=0
2*(x^2+x-20)=0
2*(x^2+x-20)/2=0/2
x^2+x-20=0
x^2-4x+5x-20=0
x(x-4)+5(x-4)=0
(x-4)(x+5)=0
x-4=0 or x+5=o
x=4 or x=-5
One number = x = 4
Other number = x+1 = 4+1 = 5
4^2+5^2=41
16+25=41
41=41