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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