Tricky solution set.

• Jan 28th 2009, 06:46 PM
algebra_fan
Tricky solution set.
The solution set for the equation: x - 4 - square root of(9x) = 0 consists of:

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

• Jan 28th 2009, 07:06 PM
CaitSydney22
are you sure?
Are you positive that there is only one positive answer? because I got two positive answers. x=16 and x=1
• Jan 28th 2009, 07:12 PM
algebra_fan
well...
It says that question 25.'s (this question) answer is B.

So I'm pretty sure... but could you show me what you did? Because I didn't even get that close to an answer.

Edit: Nevermind I get what you did! 1 doesn't work because 1-4-3= -6 not 0

so can you show me how you got 16 ?
• Jan 28th 2009, 07:16 PM
CaitSydney22
solution?
Sure,

x - 4 - sqrt(9x) = 0 add sqrt(9x) to both sides
x - 4 = sqrt(9x) now square both sides
x^2 - 8x + 16 = 9x subtract 9x from both sides
x^2 - 17x + 16 = 0 factor the left hand side
(x - 16) (x - 1) = 0 set both values equal to 0
x = 16 x = 1
• Jan 28th 2009, 07:19 PM
CaitSydney22
woops
I didn't even remember to check the answers. good job, there's your answer. 16 works, but not 1
• Jan 28th 2009, 07:24 PM
algebra_fan
Thanks!!!
I forgot one of the most basic things in Algebra, gosh I feel dumb.

Thank you!! :)