Here is a nice problem which gets many people. Not because it is tricky nor difficult.

Solve for

.

I think it gets people because they think you can solve it with basic algebraic moves.

Printable View

- January 7th 2006, 01:04 PMThePerfectHackerSolve THIS!
Here is a nice problem which gets many people. Not because it is tricky nor difficult.

Solve for

.

I think it gets people because they think you can solve it with basic algebraic moves. - January 9th 2006, 06:51 PMTreadstone 71
x=0, x=-1

You actually can solve it algebraically.

Assume x 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. - January 9th 2006, 08:14 PMThePerfectHackerQuote:

Originally Posted by**Treadstone 71**

- January 10th 2006, 03:55 AMTD!
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 - January 10th 2006, 10:08 AMTreadstone 71
You didn't specify to solve for ALL x, but to find a solution for the equation. I found two solutions.

- January 10th 2006, 10:12 AMTD!
'Solve for x' is generally used when you're asked to find all x which satisfy the equation or inequality.

- January 10th 2006, 12:07 PMThePerfectHackerQuote:

Originally Posted by**Treadstone 71**

Anyways, I happen to like this equation because it solutions are not finite. Good job, TD! - January 10th 2006, 03:28 PMThePerfectHacker

The method I used is this:

If then .

Thus, thus, one solution.

If thus, then . Thus, thus, but thus no solutions.

If thus, but this is always true! Thus, are all solutions.

Now take the union of all the solution sets

Another, non-rigorous, way of solving this problem is with a graph. Graph and graph and see where the graphs intersect. You will see the graphs coincide on - January 10th 2006, 04:14 PMTreadstone 71Quote:

Originally Posted by**TD!**