Solve THIS!

• Jan 7th 2006, 12:04 PM
ThePerfectHacker
Solve THIS!
Here is a nice problem which gets many people. Not because it is tricky nor difficult.
Solve for $x$
$|x|+|x+1|=1$.
I think it gets people because they think you can solve it with basic algebraic moves.
• Jan 9th 2006, 05:51 PM
x=0, x=-1

You actually can solve it algebraically.

Assume x $\geq$0, |x|+|x+1|=2x+1=1 => 2x=0 => x=0.

Assume x<0, |x|+|x+1|=-x+(-x-1)=-2x-1=1=>-2x=2 => x=-1.
• Jan 9th 2006, 07:14 PM
ThePerfectHacker
Quote:

x=0, x=-1

You actually can solve it algebraically.

Assume x $\geq$0, |x|+|x+1|=2x+1=1 => 2x=0 => x=0.

Assume x<0, |x|+|x+1|=-x+(-x-1)=-2x-1=1=>-2x=2 => x=-1.

HA HA got you!! What about $x=-1/2$?
• Jan 10th 2006, 02:55 AM
TD!
It can be done algebraically.

|x| + |x+1| = 1
(|x|+|x+1|)² = 1² (both sides are non-negative so squaring is allowed)
2|x(x + 1)| + 2x² + 2x + 1 = 1
2|x(x + 1)| = -2x²-2x
|x(x + 1)| = -x(x+1)
(|x(x + 1)|)² = (-x(x+1))² (*)
(x(x+1))² = (x(x+1))²

The last line is trivially true for all x, but there's an extra condition in (*).
You can only square if both sides have the same sign, the LHS is non-negative so the RHS has to be non-negative either. This gives the condition:

(x(x+1)) ≥ 0

With solution, after basic algebra: -1 ≤ x ≤ 0
• Jan 10th 2006, 09:08 AM
You didn't specify to solve for ALL x, but to find a solution for the equation. I found two solutions.
• Jan 10th 2006, 09:12 AM
TD!
'Solve for x' is generally used when you're asked to find all x which satisfy the equation or inequality.
• Jan 10th 2006, 11:07 AM
ThePerfectHacker
Quote:

You didn't specify to solve for ALL x, but to find a solution for the equation. I found two solutions.

Come on, Treadstone stop manipulating me! You knew well that you need to solve for all x. You just made a mistake and it happens. I also made many mistakes on these forums read through them and you will find them.

Anyways, I happen to like this equation because it solutions are not finite. Good job, TD!
• Jan 10th 2006, 02:28 PM
ThePerfectHacker
$|x+1|+|x|=1$
The method I used is this:

If $x\geq0$ then $x+1\geq0$.
Thus, $x+1+x=1$ thus, $x=0$ one solution.

If $x<-1$ thus, $x+1<0$ then $x<0$. Thus, $-x-1-x=-1$ thus, $x=0$ but $x<-1$ thus no solutions.

If $-1\leq x<0$ thus, $x+1-x=1$ but this is always true! Thus, $-1\leq x<0$ are all solutions.

Now take the union of all the solution sets $-1\leq x\leq 0$

Another, non-rigorous, way of solving this problem is with a graph. Graph $y=1-|x|$ and graph $y=|x+1|$ and see where the graphs intersect. You will see the graphs coincide on $-1\leq x\leq 0$
• Jan 10th 2006, 03:14 PM