1. ## Solve for x...

Alright, I just got some ridiculously long five page answer that ended up with imaginary numbers as the answer.

I think the problem is a bit more simple than what I used that's usually a dependable program to figure out an answer...

$\sqrt{x^2-4x+9}-x=-1$
$\sqrt{x^2-4x+9}-x+x=-1+x$
$\sqrt{x^2-4x+9}=0$

Get rid of the square root sign and voila...

$x-2+3=0$
$x+1=0$
$x-1=0-1$
$x=-1$

Which is the correct answer. Was my method unorthodox or a coincidence or what?

Someone help?

2. Originally Posted by A Beautiful Mind
Alright, I just got some ridiculously long five page answer that ended up with imaginary numbers as the answer.

I think the problem is a bit more simple than what I used that's usually a dependable program to figure out an answer...

$\sqrt{x^2-4x+9-x}=-1$
$\sqrt{x^2-4x+9-x+x}=-1+x$
$\sqrt{x^2-4x+9}=0$

Get rid of the square root sign and voila...

$x-2+3=0$
$x+1=0$
$x-1=0-1$
$x=-1$

Which is the correct answer. Was my method unorthodox or a coincidence or what?

Someone help?
please state the original problem from the beginning.

3. The original problem is at the top of the first post.

4. is that last $x$ supposed to be under the radical?

in other words, is the equation supposed to be

$\sqrt{x^2-4x+9} - x = -1$

because, as you have it written, $\sqrt{x^2-4x+9-x} = -1$ has no solution.

5. I just checked and no, the x is not under the radical. Sorry about that, I thought the latex stuff just covered the first 3 but I guess not.

6. $\sqrt{x^2-4x+9} - x = -1$

$\sqrt{x^2-4x+9} = x - 1$

square both sides ...

$x^2 - 4x + 9 = (x-1)^2$

$x^2 - 4x + 9 = x^2 - 2x + 1
$

combine like terms ...

$8 = 2x$

$x = 4$

check the solution in the original equation ...

$\sqrt{4^2-4(4)+9} - 4 =$

$\sqrt{16-16+9} - 4 =$

3 - 4 = -1

checks good.