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Question 295131: 6) Find 3 consecutive intergers such that the sum of their squares is 50.
7) Find 3 consecutive even intergers such that the sum of the squares of the first two less the square of the third equals 20.
8) Find 3 consecutive odd intergers such that the sum of the squares of the second and third exceeds the square of the first by 209.
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! let the integers be x , x+1, x+2
x^2+(x+1)^2 +(x+2)^2=50
x^2+x^2+2x+1 + x^2+4x+4 =50
3x^2+6x+5=50
3x^2+6x-45=0
x^2+2x-15=0
x^2+5x-3x-15=0
x(x+5)-3(x+5)=0
(x+5)(x-3)=0
x=-5 or x=3
the numbers can be -5, -4, -3 OR 3 , 4 , 5
Both the solutions satisfy the equation
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