# Solving for x

• Jan 27th 2009, 06:03 PM
DCHomage
Solving for x
Solve for x: x^2+1=x
• Jan 27th 2009, 06:06 PM
Jhevon
Quote:

Originally Posted by DCHomage
Solve for x: x^2+1=x

this is a quadratic equation, do you know how to deal with those?

first note that you have $\displaystyle x^2 - x + 1 = 0$

next you would try to factor, but it won't work here, so the next move is to complete the square, or use the quadratic formula. can you continue?
• Jan 27th 2009, 06:11 PM
DCHomage
Let's see. Do I move the -x+1 over?
• Jan 27th 2009, 06:31 PM
mr fantastic
Quote:

Originally Posted by DCHomage
Let's see. Do I move the -x+1 over?

Quote:

Originally Posted by Jhevon
this is a quadratic equation, do you know how to deal with those?

first note that you have $\displaystyle x^2 - x + 1 = 0$

next you would try to factor, but it won't work here, so the next move is to complete the square, or use the quadratic formula. can you continue?

Do you know what completing the square means? Are you familiar with the quadratic formula?
• Jan 27th 2009, 06:32 PM
Jhevon
Quote:

Originally Posted by DCHomage
Let's see. Do I move the -x+1 over?

were that the case, i would have done it in the first step.

the next move, i said, is to complete the square, or use the quadratic formula. are you familiar with these methods?
• Jan 27th 2009, 06:57 PM
DCHomage
can't remember. I have been working at my full time job that i forgot about it.
• Jan 27th 2009, 07:05 PM
Jhevon
Quote:

Originally Posted by DCHomage
can't remember. I have been working at my full time job that i forgot about it.

the easier way would be the quadratic formula. it says the following

the solution(s) to an equation of the form $\displaystyle ax^2 + bx + c = 0$ are given by $\displaystyle x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$

what would that give as the answer to this problem?
• Jan 27th 2009, 07:12 PM
DCHomage
Not sure. I have this so far
$\displaystyle x = \frac {x \pm \sqrt{x^2 - 4x^2}}{2x^2}$
• Jan 27th 2009, 07:24 PM
mr fantastic
Quote:

Originally Posted by DCHomage
Not sure. I have this so far
$\displaystyle x = \frac {x \pm \sqrt{x^2 - 4x^2}}{2x^2}$

No.

Compare $\displaystyle x^2 - x + 1$ with $\displaystyle ax^2 + bx + c$ and it should be clear that a = 1, b = -1 and c = 1. These are the values that get substituted into the quadratic formula.
• Jan 27th 2009, 07:39 PM
DCHomage
I got this so far.

$\displaystyle x = \frac {1 \pm \sqrt{-3}}{2}$
• Jan 27th 2009, 07:42 PM
mr fantastic
Quote:

Originally Posted by DCHomage
I got this so far.

$\displaystyle x = \frac {1 \pm \sqrt{-3}}{2}$

Yes. So there's no real solution. (There are however two non-real solutions).
• Jan 27th 2009, 07:48 PM
DCHomage
So, I can't solve for x?
• Jan 27th 2009, 07:52 PM
mr fantastic
Quote:

Originally Posted by DCHomage
So, I can't solve for x?

Quote:

Originally Posted by mr fantastic
[snip]
there's no real solution. (There are however two non-real solutions).

..