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About LinkBacksHere is a nice problem which gets many people. Not because it is tricky nor difficult.
Solve for $\displaystyle x$
$\displaystyle |x|+|x+1|=1$.
I think it gets people because they think you can solve it with basic algebraic moves.
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
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.Originally Posted by Treadstone 71
Anyways, I happen to like this equation because it solutions are not finite. Good job, TD!
$\displaystyle |x+1|+|x|=1$
The method I used is this:
If $\displaystyle x\geq0$ then $\displaystyle x+1\geq0$.
Thus, $\displaystyle x+1+x=1$ thus, $\displaystyle x=0$ one solution.
If $\displaystyle x<-1$ thus,$\displaystyle x+1<0$ then $\displaystyle x<0$. Thus, $\displaystyle -x-1-x=-1$ thus, $\displaystyle x=0$ but $\displaystyle x<-1$ thus no solutions.
If $\displaystyle -1\leq x<0$ thus, $\displaystyle x+1-x=1$ but this is always true! Thus, $\displaystyle -1\leq x<0$ are all solutions.
Now take the union of all the solution sets $\displaystyle -1\leq x\leq 0$
Another, non-rigorous, way of solving this problem is with a graph. Graph $\displaystyle y=1-|x|$ and graph $\displaystyle y=|x+1|$ and see where the graphs intersect. You will see the graphs coincide on $\displaystyle -1\leq x\leq 0$
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