SOLUTION: The product of two consecutive integers is 90 less than twice the square of the next larger integer. Find the smaller integer. A. 6 B. 8 C. 10 D. 12

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Question 1198815: The product of two consecutive integers is 90 less than twice the square of the next larger integer. Find the smaller integer.
A. 6
B. 8
C. 10
D. 12
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3835)   (Show Source): You can put this solution on YOUR website!

The first sentence translates to this equation
n(n+1) = 2(n+1)^2-90

n = first integer
n+1 = integer right after n

Let's solve for n.
n(n+1) = 2(n+1)^2-90
n^2+n = 2(n^2+2n+1)-90
n^2+n = 2n^2+4n+2-90
n^2+n = 2n^2+4n-88
0 = 2n^2+4n-88-n^2-n
n^2+3n-88 = 0
(n+11)(n-8) = 0
n+11 = 0 or n-8 = 0
n = -11 or n = 8

If we focus on the positive values only, then n = -11 is ignored.

Answer: Choice B
Answer by ikleyn(53763)   (Show Source): You can put this solution on YOUR website!
.

Your wording in this problem is  FATALLY  incorrect and leads to confusing.

The correct wording is as follows 

    The product of two consecutive integers is 90 less than twice 
    the square of the  larger of the two integers. 
    Find the smaller integer.



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