x- 4 - sq. rt. Of: 9x = 0 consists of



A)exactly one negative number
B) exactly one positive number - B
C) exactly one positive number and one negative number.
D) exactly two negative numbers
E) exactly two positive numbers

O.K, someone else was able to help me with this, I forgot something minor and messed up the whole problem (feel like an idiot :D ) .

Here's the solution for anyone who's curious:

x-4- sq.rt.(9x)=0 add sq.rt.(9x) to both sides

x-4=sq.rt.(9x) square both sides

(x-4)^2= 9x expand

x^2 - 8x + 16 = 9x subtract 9x

x^2 - 17x + 16 = 0 factor


x=16 and x=1 plug in both numbers, 1 doesn't work
so x is simply 16.