a 3 consecutive positive integer are such that the sum of the square of the smallest number and the product of the other 2 is 46 find the integer
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The integers are 4,5,6.
Let's consider the three consecutive integers as x, x+1, x+2.
Now if we try mapping the condition, we'll get
x² + (x + 1) * (x + 2) = 46
x² + x² + 2x + 1x + 2 = 46
2x² + 3x = 44
2x² + 3x - 44 = 0
After solving the equation we'll get,
2x + 11 = 0 and x - 4 = 0
We're looking for positive integers so our x will be 4.
Thus the other two integers will be 5, 6.
Let's consider the three consecutive integers as x, x+1, x+2.
Now if we try mapping the condition, we'll get
x² + (x + 1) * (x + 2) = 46
x² + x² + 2x + 1x + 2 = 46
2x² + 3x = 44
2x² + 3x - 44 = 0
After solving the equation we'll get,
2x + 11 = 0 and x - 4 = 0
We're looking for positive integers so our x will be 4.
Thus the other two integers will be 5, 6.
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