# Absolute value equality solution

#### marksyncm

1. The problem statement, all variables and given/known data

Find solutions to the given equality

2. Relevant equation

$$x^2 +3|x-1|=1$$

3. The attempt at a solution

The above can be rewritten as:

$$|x-1| = {1-x^2\over 3}$$

If my understanding of absolute values is correct, the above simply means that:

$$x-1 = {1-x^2\over 3} \lor x-1 = -({1-x^2\over 3})$$

For the first equation:

$$x-1 = {1-x^2\over 3} = 3x-3=1-x^2 \iff x^2+3x-4=0 \iff x \in \{-4, 1\}$$

For the second equation:

$$x-1 = -({1-x^2\over 3}) \iff 1-x={1-x^2\over 3} \iff x^2-3x+2 =0 \iff x \in \{1, 2\}$$

So apparently, any $x \in \{-4, 1, 2\}$ fulfills at least one of the $\lor$ stipulations for the absolute value, so I would expect either of these three values to fulfill the original equation. However, only $x=1$ is correct - the intersection of the solutions to the two absolute value equations.

What am I missing?

As an example, if we were solving $|x-1|=5$, this would mean that either $x-1=5 \lor x-1=-5$. This means that $x=6 \lor x=-4$ and I expect either of these solutions to solve the original equation, not just the intersection of the two solutions, which in this case would be $\emptyset$.

I'm probably missing something extremely obvious here.

EDIT:

Is this because, in the original equation $|x-1| = {1-x^2\over 3}$, the domain is restricted to values where ${1-x^2\over 3} \geq 0$? This would restrict the domain to $-1 \geq x \leq 1$, which explains the result.

Thanks

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#### symbolipoint

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Look at two possibilities for x-1.
x-1<0, and OR x-1>=0.
Now, solve for EACH condition separately.

You will solve this system:
x^2+3(-x+1)=1
and
x^2+3(x-1)=1.

EDIT: I am changing one word, hopefully to make the conditions a little clearer.

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#### marksyncm

Thanks. Why are we interested in whether x-1 is positive or negative when what comes "out" of the absolute value symbols is always positive? Ie. isn't the ultimate value of the equation determined by |x-1|, which is always positive, rather than by the sign of what's inside the absolute value pipes, which is x-1? Shouldn't we be more concerned with the sign of terms that are outside of absolute value pipes? ie. if $|x+5| = x-2$, the right side of the equation needs to be $\geq 0$, otherwise an equality is impossible.

Or are these just two sides of the same coin?

#### verty

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What is the solution when $x \geq 1$?

#### symbolipoint

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Thanks. Why are we interested in whether x-1 is positive or negative when what comes "out" of the absolute value symbols is always positive? Ie. isn't the ultimate value of the equation determined by |x-1|, which is always positive, rather than by the sign of what's inside the absolute value pipes, which is x-1? Shouldn't we be more concerned with the sign of terms that are outside of absolute value pipes? ie. if $|x+5| = x-2$, the right side of the equation needs to be $\geq 0$, otherwise an equality is impossible.

Or are these just two sides of the same coin?
Sign of what's inside the absolute value function, very important. Be reminded, the expression inside may often be either non-negative or negative; while the absolute value is never negative. This is why we must examine two cases for one absolute value expression.

#### symbolipoint

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What is the solution when $x \geq 1$?
Think that through!

Your is in part of your original equation, |x-1|.
One case is for x-1>=0, which will mean x>=1.

The other case is for x-1<0, which will mean x<1.

Look again at the resulting system of equations to solve. Note that x at 1 is a critical x value. The original equation has a particular result for x<1 and a different result for x>=1.

#### symbolipoint

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I now solved this equation. If the bosses approve of the posting, I will post what I found, which some members may scrutinize for any needed remarks for better solution.

#### marksyncm

What is the solution when $x \geq 1$?
If $x=1$ then the equation holds ($0 = 0$). If $x > 1$ then the equation is invalid because the right side of $|x-1| = {1-x^2\over 3}$ would be negative, which is a contradiction because the left side is always positive or equal to zero, so the right side must always be positive or equal to zero as well. So the domain of $|x-1| = {1-x^2\over 3}$ is restricted to $x \in <-1,1>$ because only for those values are both sides of the equation positive or equal to zero. We can therefore disregard any solutions that fall outside this range. Can you point out the flaw in this reasoning?

I now solved this equation. If the bosses approve of the posting, I will post what I found, which some members may scrutinize for any needed remarks for better solution.
Just to be clear, the solution to the equation is already presented in the original post. It is $x=1$.

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• Delta2

#### verty

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Can you point out a flaw in this reasoning?
Yes, you made it impossible for the answer to be 1.

Back now to the topic. Do you understand Symbolipoint's method of solving it? That is my preferred method as well.

PS. I'll leave you and him to talk now.

#### marksyncm

Yes, you made it impossible for the answer to be 1.
I fixed my original statement. Here's how it reads now:

"If $x=1$ then the equation holds ($0 = 0$). If $x > 1$ then the equation is invalid because the right side of $|x-1| = {1-x^2\over 3}$ would be negative, which is a contradiction because the left side is always positive or equal to zero, so the right side must always be positive or equal to zero as well."

Does this make sense now?

Back now to the topic. Do you understand Symbolipoint's method of solving it?
I don't, unfortunately. I understand the motions and can follow the instructions, but I do not understand it on an intuitive level.

How about this: is my method (after all corrections) OK, or is there something wrong with it? I might be able to make more sense of Symboilpoint's method if I attempt to "convert" my own method to it, but I can only attempt that if I'm certain my method is correct.

#### verty

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Well -4 is discounted and so is 2. If you get that, you understand the method, and regardless, I'm not going to explain anything further. Practice more of these.

PS. You method is fine and it's your method so use it unless it fails.

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#### symbolipoint

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Back to original post #1, the first step, |x-1|=(1-x^2)/3, does not help. Also it is not needed.

#### symbolipoint

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Well -4 is discounted and so is 2. If you get that, you understand the method, and regardless, I'm not going to explain anything further. Practice more of these.
One should check those values in the original equation to find if they make sense or not.

#### marksyncm

I gave this some more thought yesterday and finally managed to find an answer to my question that I personally find intuitive and which addresses my concern. I'm posting this for others who might face a similar problem, as well as to consolidate my own thoughts (and to hopefully get feedback if anything sounds off).

I'll use the following to illustrate:

$$|x+1|=2x$$

Here's how such equations are typically solved:

$$x+1 \geq 0 \iff x \geq -1$$
$$x+1 < 0 \iff x < -1$$

If $x+1$ is non-negative, then $x+1=2x \iff x=1$. Because $1 \geq -1$ (ie. it is within the set of solutions that make $x+1 \geq 0$), it is a solution to the original equation.

Conversely, if $x+1$ is negative, then $x+1=-(2x) \iff -3x=1 \iff x=-\frac{1}{3}$. Because $-\frac{1}{3}$ is not<-1 (ie. it is not within the set of solutions that make $x+1 <0$), it is not a solution to the original equation

This wasn't intuitive for me because I failed to grasp why this worked. I understood the steps, but I didn't understand why following this solving method lead to the correct solution. Examples of confusing questions that popped into my head included:"

"What if, when solving for $x+1=-(2x)$, the solution turned out to be $< -1$ instead of $-\frac{1}{3}$? Was this for some reason algebraically impossible? And if it did turn out to be $< -1$, would that then mean that $x$ can be a negative number? This would mean that $2x$ in the original equation was a negative number, and a negative number cannot be equal to an absolute value..."

Basically, I did not understand why discarding or keeping solutions based on whether they fell inside the appropriate range of solutions allowed us to find perfect solutions for absolute value equations of this kind. I understood why it worked algebraically, but I did not get it intuitively.

Here's the insight that cleared up my confusion:

What the above restrictions on the value of $x$ do is that they guarantee that $2x$ (ie. the right side of the original equation) is never negative. Going back to the original equation and if solving for the case where $x+1$ is non-negative, we have:

$$x+1=2x$$

Why would we discard any solution that's not $\geq -1$? Because any solution outside this range would make $x+1$ negative, which would in turn make $2x$ negative - which is impossible because $2x$ is supposed to be equal to $|x+1|$, and a negative number cannot be equal to an absolute value.

Similarly, when solving for the case where $x+1$ is negative, we have:

$$x+1=-(2x)$$

Why would we discard any solution that's not $<-1$? Because any solution outside this range would make $x+1$ positive, which would mean that $-(2x)$ is positive, which would mean that $2x$ is negative - which is impossible for the same reason outlined above.

I realize this may look like I'm confusing matters more than I'm helping them, but that's the sort of understanding that I was looking for. Hopefully someone else finds it useful.

#### verty

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Well you make an assumption, so you can't forget about the assumption. If you assume, to make progress, that $x \geq 1$, then a candidate under that assumption can't be a solution if it goes against the assumption. Is it solving by cases or by assumptions, I don't know. But, assumptions are very important.

#### symbolipoint

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Just a reminder:
I now solved this equation. If the bosses approve of the posting, I will post what I found, which some members may scrutinize for any needed remarks for better solution.
Sign of what's inside the absolute value function, very important. Be reminded, the expression inside may often be either non-negative or negative; while the absolute value is never negative. This is why we must examine two cases for one absolute value expression.

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#### symbolipoint

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Original Equation
x2+3|x−1|=1
|x-1| must be non-negative.
x-1 may be non-negative or negative.

EITHER
x2+3(x-1)=1
x2+3x-3=1
x2+3x-4=0
and then
(x-4)(x+1)=0
which means
x=-1 or x=4

OR
x2+3(-x+1)=1
x2-3x+3=1
x2-3x+2=0
and then
(x-2)(x-1)=0
which means
x=1 or x=2

Check to be sure that each of these does or does not satisfy the original equation:
x=-1, or x=1, or x=2, or x=4

#### Ray Vickson

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Thanks. Why are we interested in whether x-1 is positive or negative when what comes "out" of the absolute value symbols is always positive? Ie. isn't the ultimate value of the equation determined by |x-1|, which is always positive, rather than by the sign of what's inside the absolute value pipes, which is x-1? Shouldn't we be more concerned with the sign of terms that are outside of absolute value pipes? ie. if $|x+5| = x-2$, the right side of the equation needs to be $\geq 0$, otherwise an equality is impossible.

Or are these just two sides of the same coin?
When we write $|a|$ we are definitely interested in whether or not $a \geq 0$, because the definition of the absolute value function is
$$|a| = \begin{cases} a & \text{if} \; a \geq 0\\ -a & \text{if} \; a < 0 \end{cases}$$
So, $|a| \geq 0$, no matter what sign we choose for $a$, but whether or not we must use $+a$ or $-a$ in an equation containing $|a|$ is of crucial importance

• symbolipoint

#### symbolipoint

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When we write $|a|$ we are definitely interested in whether or not $a \geq 0$, because the definition of the absolute value function is
$$|a| = \begin{cases} a & \text{if} \; a \geq 0\\ -a & \text{if} \; a < 0 \end{cases}$$
So, $|a| \geq 0$, no matter what sign we choose for $a$, but whether or not we must use $+a$ or $-a$ in an equation containing $|a|$ is of crucial importance
That helps in explaining the logic. Also showing this definition using a number line helps in explaining the definition.

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