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Question
The sum of the squares of two numbers as 233 and one of the numbers as 3 less than twice the other number find the numbers.
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Solution
Let the numbers be integers. One of the numbers be x. So, the other will be (2x - 3).
Then according to question,
x2 + (2x - 3)2 = 233
x2 + 4x2 - 12x + 9 = 233
5x2 - 12x + 9 - 233 = 0
5x2 - 12x - 224 = 0
5x2 - 40x + 28x - 224 = 0
5x(x - 8) + 28(x - 8) = 0
(x - 8)(5x + 28) = 0
x - 8 = 0
x = 8
Or
5x + 28 = 0
5x = -28
x = -28/5
Since, we have assumed the numbers to be integers, so x cannot be a rational number/fraction.
Therefore, for x = 8
Other number = (2x - 3) = 2(8) - 3 = 16 - 3 = 13
Thus, whole numbers be 8, 13.
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