SOLUTION: The sum of the squares of two consecutive integers is 9 greater than 8 times the smaller integer. Find the integers.

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Question 391464: The sum of the squares of two consecutive integers is 9 greater than 8 times the smaller integer. Find the integers.
Answer by MathLover1(20850) (Show Source):
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two consecutive integers are n, n +1

The sum of the squares of two consecutive integers is

........solve for

Solved by pluggable solver: Quadratic Formula

Let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=1, b=-3, and c=-4

Negate -3 to get 3

Square -3 to get 9 (note: remember when you square -3, you must square the negative as well. This is because .)

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

or

Lets look at the first part:

Add the terms in the numerator

Divide

So one answer is

Now lets look at the second part:

Subtract the terms in the numerator

Divide

So another answer is

So our solutions are:

or

so, you have
and
second integer
4 and 5

second integer
-1 and 0
check: 4 and 5

other pair: -1 and 0