SOLUTION: one number is 1 more then 3 times another. If the sum of their squares is 689, find the numbers: (AC method to solve, use whole numbers only)
So far I have:
x(3x+1)=689
3x sq
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Question 462309: one number is 1 more then 3 times another. If the sum of their squares is 689, find the numbers: (AC method to solve, use whole numbers only)
So far I have:
x(3x+1)=689
3x squared + 1x = 689
3x squared + 1x +
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
"sum of their squares" ___ x^2 + (3x + 1)^2 = 689 ___ x^2 + 9x^2 + 6x + 1 = 689
10x^2 + 6x - 688 = 0 ___ 5x^2 + 3x - 344 = 0
factoring ___ (5x + 43)(x - 8) = 0
5x + 43 = 0 ___ x = -43/5 ___ not a whole number
x - 8 = 0 ___ x = 8
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