SOLUTION: Find two consecutive odd integers such that the square of the larger one is 25 more than 8 times the smaller one.

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Question 226107: Find two consecutive odd integers such that the square of the larger one is 25 more than 8 times the smaller one.
Found 2 solutions by solver91311, imurbabiigurl:
Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

Let represent the smaller odd integer. Then must represent the next consecutive odd integer.

The square of the larger one () is (=) 25 more than 8 times the smaller one ()

Solve the quadratic for to get the value of the smaller integer. It factors as one would presume since you are expecting integer answers. You will get two roots. Check to see if both of them are valid answers remembering that integers can be negative, but that the next consecutive integer to a negative integer has a smaller absolute value.

John

Answer by imurbabiigurl(1) (Show Source):
You can put this solution on YOUR website!
Find two consecutive odd integers such that the square of the larger one is 25 more than 8 times the smaller one.
= the smaller odd integer
= the next consecutive integer

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

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=100 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 7, -3. Here's your graph:

Staci :]