solve the following $|x-1| + |x+1|= x-3$

Question

I guess it is simple but I am trying it two different ways and I am not getting the same answer Donno why $$|x-1|+|x-1|=x-3$$ Can somebody solve it by squaring both sides

Answer 1

See below the graphical representations that ILoveMath mentions.

The $U$ shape curve is for $y=|x-1|+|x+1|$, the straight line is for $y=x-3.$

No intersections between the curves, meaning no solution in $(x,y)$; therefore no solution for $x$.

For a rigorous treatment, a systematic method (without subtleties, I admit) transforms the given equation into the three following ones, none of them having a solution in the range of values of $x$ for which they are valid:

$$\begin{cases}\text{If} & x<-1 : &-(x-1)-(x+1)=x-3 & \text{sol:} \ x=1 & \text{not compatible with} \ x<-1 \\ \text{If} & -1 \leq x \leq 1 : & -(x-1)+(x+1)=x-3 & \text{sol:} \ x=5 & \text{not compatible with} \ -1 \leq x \leq 1 \\ \text{If} & x> 1 : & (x-1)+(x+1)=x-3 : & \text{sol:} \ x=-3 & \text{not compatible with} \ x>1 \end{cases}$$

Remark: spurious solutions like $x=1$, $x=5$ and $x=-3$ can be explained: for example pseudo-solution $x=1$ is nothing else than the abscissa of intersection point (denoted $A$ on the graphics) of the prolongation (dotted line) of $y=-(x-1)-(x+1)$, i.e., $y=-2x$ with $y=x-3$.

Answer 2

For all $x\in {\Bbb R}$ we have: $$ x-3< x \leq |x| \leq 2|x| = |x+1+x-1| \leq |x+1|+|x-1|$$ so there are no solutions.

Answer 3

The system has $\mathbf{NO}$ solutions. Let $f(x) = |x-1|+|x+1|$.

Notice, if $|x|>1$, then you get either $f(x) = 2x $ or $f(x) = -2x$ while if $|x|<1$, then $f(x) = 2$

Now, do the graphs and see that there is indeed no solutions.

Answer 4

Note that the left hand side is never negative. If $x<3$, then the right hand side is negative. If $x\ge 3$, then the right hand side is positive, $x-1$ is positive, and $x+1$ is positive. What does this tell you about the absolute values? Are there any solutions?

Answer 5

You might try and solve the equation by repeated squaring, but there are some caveats, before doing so.

You can't square both sides of an equation unless you are sure that both sides are nonnegative or nonpositive, otherwise you could introduce spurious solutions.

In your case, the left-hand side is clearly nonnegative (actually positive), so the right-hand side must be nonnegative as well. This implies $x\ge3$. But now squaring is not necessary, because if $x\ge3$ then $x\ge1$ and $x\ge-1$; thus the equation becomes $$ \begin{cases} x\ge 3 \\[4px] x-1+x+1=x-3 \end{cases} $$ The equation has solution $x=-3$, which doesn't satisfy the inequality above, so we conclude that your equation has no solution.

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What if we square carelessly? Let's try. Write the equation as $$ |x-1|=x-3-|x+1| $$ and square, getting $$ x^2-2x+1=x^2-6x+9+x^2+2x+1-2(x-3)|x+1| $$ Simplify and isolate the absolute value: $$ 2(x-3)|x+1|=x^2-2x+9 $$ Square again: $$ 4(x-3)^2(x-1)^2=(x^2-2x+9)^2 $$ that becomes $$ 4x^4 - 16x^3 - 8x^2 + 48x + 36 = x^4 - 4x^3 + 22x^2 - 36x + 81 $$ or $$ 3x^4 - 12x^3 - 30x^2 + 84x - 45 = 0 $$ that factors as $$ 3(x-1)^2(x-5)(x+3)=0 $$ One might think that the solutions are $1$, $5$ and $-3$, but substituting them in the original equation shows they aren't. For instance, substituting $x=5$ gives the false equality $$ |5-1|+|5+1|=5-3 $$ and you can check the other two values.

This is because squaring has introduced spurious solutions. Such a procedure only gives necessary conditions for the roots, but in general they are not sufficient. After all, $1^2=(-1)^2$, but we cannot conclude $1=-1$.

Some intelligence work before starting a mechanical procedure helps in doing less computations.

Answer 6

(a)If $x \leq -1$, then $$\vert x - 1 \vert + \vert x + 1 \vert= - (x-1) - (x+1) = -2x$$Equate it to $-2x = x - 3$, which gives $$x = 1$$ Therefore, no solution

(b)If $-1 \leq x \leq 1$, then $$\vert x - 1 \vert + \vert x + 1 \vert = - (x-1) + (x+1) = 2$$Equate it to $2 = x - 3$, which gives $$x = 5$$ Therefore, no solution

(c)If $x \geq 1$, then $$\vert x - 1 \vert + \vert x + 1 \vert = (x-1) + (x+1) = 2x$$Equate it to $2x = x - 3$, which gives $$x = -3$$ Therefore, no solution