SOLUTION: "find two consecutive positive integers such that the square of the first is decreased by 17 equals 4 time the second."

Question 895789: "find two consecutive positive integers such that the square of the first is decreased by 17 equals 4 time the second."
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
x is the first number.
x+1 is the second number.

your equation becomes:

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

simplify this equation to get:

x^2 - 17 = 4x + 4

subtract 4x and subtract 4 from both sides of this equation to get:

x^2 - 17 - 4x - 4 = 0

simplify to get:

x^2 - 4x - 21 = 0

factor to get:

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

solve for x to get:

x = 7 or x = -3

x has to be a positive integer so your solution appears to be 7.

first number is 7
second number is 8

7^2 - 17 = 49 - 17 = 32
4*8 = 32
first number squared minus 17 is equal to 4 times the second number so your solution is good.

solution is x = 7.

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