SOLUTION: There are two integers. The square of the first integer added to ten times the second integer is equal to -9. What are the integers?

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Question 1150440: There are two integers. The square of the first integer added to ten times the second integer is equal to -9. What are the integers?
Found 2 solutions by Velo, ikleyn:
Answer by Velo(3) About Me  (Show Source):
You can put this solution on YOUR website!

1 and -1 will be your integers.

Remember, integers are whole numbers that are negative, positive, or zero.

Therefore, let the first integer be x%5E2 and the second integer 10%28-x%29 since your answer will be a negative number, -9.

We can set up the equation:

x%5E2+%2B+10%28-x%29+=+-9

1%5E2+%2B+10%28-1%29+=+-9

1+-+10+=+-9

Answer by ikleyn(52802) About Me  (Show Source):
You can put this solution on YOUR website!
.

The setup in the post by  @Velo is  INCORRECT.

The correct setup is  THIS  equation


            x^2 + 10y = -9,         (1)


where  "x"  is the first integer number and  "y"  is the second integer number.

Equation  (1)  has  INFINITELY  MANY  solutions in integer numbers.


The first several solutions are as follows

        (x,y) = (1,-1), (9,-9), (11,-13), (19.-37), (21,-45), (29,-85), (31,-97).