Find two consecutive positive integers such that the square of the first decreased by 17 equals 4 times the second.

Let the first number = x

The second number would be x + 1  ( this makes the two numbers consecutive).

Now you want the square of the first one to decrease by 17, so now you have x^2 -17.

Then you want that to equal 4 times the second number so you now have:

x^2 -17 = 4(*x+1)

Now we can solve for x:

x^2 -17 = 4x +4

Subtract 4x+4 from both sides:

x^2 - 4x -21 = 0

Factor:

(x-7) (x+3) = 0

Solve for the two x's:

x = 7 and x = -3

The problem states the numbers are positive, so -3 cannot be an answer, so we have X = 7

The first number is X , which is 7 and the second number is x +1 which is 7+1 = 8

The two numbers are 7 and 8.

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step-by-step explanation:

1. is 4 and 2. is 3

step-by-step explanation:

so if you look at the first one you see that 1 is to 4 and 2 is to 8 and so on, so you can see for every 1 there is a times 4 so if it were 3 then it would be 3*4 and it is 12

for number 2 it is 3 because 1 is to 3 and 2 is to 6 so for every 1 there is 3 so if it were 3 then there would be 3*3 = 9