SOLUTION: When one integer is added to the square of the next consecutive integer, the sum is 41. Find the integers.
Please show how to solve this?
Algebra.Com
Question 286294: When one integer is added to the square of the next consecutive integer, the sum is 41. Find the integers.
Please show how to solve this?
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
The answer that springs to mind is 5 + 6*6 = 41.
But that tells the story without showing the work.
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x = one integer
x+1 = the next consecutive integer
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x + (x+1)^2 = 41
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Collect terms and simplify
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x + x^2 + 2x + 1 = 41
x^2 + 3x -40 = 0
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factor
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(x+8)(x-5) = 0
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So we have TWO possible answers: x= -8, x=5.
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5 and 6 is one answer, as mentioned above and confirmed.
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But by doing the work, we have another candidate.
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With quadratic terms you sometimes get what are called spurious answers: answers that cannot be correct. The answer x= -8 might not work, but let's check it.
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-8 + -7^2 = -8 + 49 = 41
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Correct! That show how important it is to do and to show the work.
.
Answer:
The two consecutive integers are 5 and 6 or -8 and -7.
.
Done.
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