Question

Find 3 consecutive odd numbers where the product of the smaller two numbers is 34 less than the square of the largest number. @alli14344

Answer 1

3,5,7

Answer 2

x > y > z
x - 2 = y
y - 2 = z
yz = x^2 - 34

x - 2 = y
x = y + 2

yz = x^2 - 34
y(y - 2) = (y + 2)^2 - 34
y^2 - 2y = (y + 2)(y + 2) - 34
y^2 - 2y = y^2 + 2y + 2y + 4 - 34
y^2 - y^2 - 2y - 4y = 4 - 34
-6y = -30
y = -30/-6
y = 5

x - 2 = y
x - 2 = 5
x = 5 + 2
x = 7

y - 2 = z
5 - 2 = z
z = 3

∴ x = 7 , y = 5 , z = 3


or you could do it this way:

suppose the smallest odd number was x
the second number would be x + 2, and the third number would be x + 4
so the product of the two smaller numbers (x, x + 2) is 34 less than the square of the largest number (x + 4)
x(x+2) = (x+4)^2 - 34
x^2 + 2x = x^2 + 8x + 16 - 34
6x = 18
x = 3
so the three number are 3, 5, and 7