SOLUTION: Find two consecutive odd integers such that the square root of the larger is four less than the smaller.

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Question 579097: Find two consecutive odd integers such that the square root of the larger is four less than the smaller.
Answer by dfrazzetto(283)   (Show Source): You can put this solution on YOUR website!
Intuitively, we can guess it is 7 and 9, which works, but to solve it mathematically
2n+1, 2n+3






Square both sides:


2n+3 = 4n^2 -12n + 9
4n^2 - 14n + 6 = 0; divide by 2


2n^2 - 7n + 3 = 0; factor:


(2n - 1)(n - 3) = 0
n = 1/2 or 3
plugging 1/2 into original equations gives us 2 and 4 which are even, so throw that out
plugging 3 into original equations gives us 7 and 9, which work
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=25 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 3, 0.5. Here's your graph:

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