# Solve for ##x## involving modulus

• brotherbobby
In summary: Yes. My point was really if we draw a graph of the function, then all is clear.It is necessary that the solutions lie in said interval.
brotherbobby
Homework Statement
Solve ##|x|+|x-2|=2##
Relevant Equations
1. For a given function ##f(x)##, if ##|f(x)|<a \Rightarrow -a < f(x) < a##
2. For a given function ##f(x)##, if ##|f(x)|>a\Rightarrow f(x)<-a\; \text{or}\; f(x)>a##
(I could solve the problem but could not make sense of the solution given in the text. Let me put my own solutions below first).

1. Problem Statement :
I copy and paste the problem to the right as it appears in the text.

2. My attempt : There are three "regions" where ##x## can lie.

(1) ##\mathbf{x \ge 2}## : In this case we have given ##|x|+|x-2|=2\Rightarrow x+x-2 = 2\Rightarrow 2x=4\Rightarrow \boxed{x=2}## (as the answer ##x=2## lies within the selected range in which the equation is solved).

(2) ##\mathbf{0\le x<2}## : For this case, the equation: ##|x|+|x-2|=2\Rightarrow x+2-x = 2 \Rightarrow 2=2## which is always true. Hence the selected range is a solution : ##\boxed{0\le x<2}##.

(3) ##\mathbf{x<0}## : In this case, the given equation : ##|x|+|x-2|=2\Rightarrow -x+2-x = 2\Rightarrow 2x=0\Rightarrow x=0##. This solution has to be rejected for it doesn't lie in the selected range.

Hence, the solution to the problem : ##\boxed{0\le x\le 2}##. ##\Large{\checkmark}##, as it agrees with the answer in the text.

3. The text's attempt :
I copy and paste the solution in the text given to the right.

It is the solution that I am struggling to follow. Namely on the two steps marked as 1 and 2.

In 1, the author is tacitly assuming that ##x>0## and that ##x<2##. What is the reason for this assumption? There are three broad regions where ##x## can lie, as I have shown in my solution above. Why take the one on the middle, as he has done?

2 is more confusing. How can the expression ##x-(x-2)=2## become ##x(x-2)\le 0##?

A hint or suggestion referring to the author's solution would be welcome. I admit that he obtains the same answer as I do.

Delta2
I wouldn't particularly worry about the text solution as yours is better.

Note that the one-way implication given in the text is not enough. It needs two-way implication.

Re your first question, the author has replaced the ##2## on the RHS by the equivalent ##x - (x -2)##.

Delta2
@PeroK said how (1) result. For the (2), just square both sides of (1). After simplifications you ll get $$|x||x-2|=-x(x-2)\iff |x(x-2)|=-x(x-2)$$

PeroK
PeroK said:
Re your first question, the author has replaced the ##2## on the RHS by the equivalent ##x - (x -2)##.
Ahh how silly of me. Yes indeed ##2 = x-(x-2)##, as an identity, independent of the problem. The author's solution not only makes sense but is superior to my own. Let me explain.

But first, I forgot to mention something in the Relevant Equations. That ##|x|+|y| = |x-y|##, if ##x## and ##y## are of opposite signs. Or one of them can be zero.

Attempt :
\begin{equation*}
\begin{split}
|x|+|x-2|&=2\\
& = |2|\\
&=|x-(x-2)|\\
\Rightarrow x(x-2)&\le 0 \mathbf{*}\\
&\Rightarrow\boxed{0\le x \le 2}
\end{split}
\end{equation*}

* Since the line above implies that (x) and (x-2) must be of opposite signs, referring to my Relevant Equations.

Delta2
A simple approach is to draw a graph of the function ##|x| + |x-a|## and note that it has a constant value of ##a## in the interval ##[0,a]## and linear on either side.

That also solves the more general problem of ##|x| + |x-a| = b##.

Delta2
PeroK said:
A simple approach is to draw a graph of the function ##|x| + |x-a|## and note that it has a constant value of ##a## in the interval ##[0,a]## and linear on either side.

That also solves the more general problem of ##|x| + |x-a| = b##.
ehm was that a typo you meant to write the more general problem ##|x|+|x-a|=a##?

Delta2 said:
ehm was that a typo you meant to write the more general problem ##|x|+|x-a|=a##?
No, the general case where ##a \ne b##.

PeroK said:
No, the general case where ##a \ne b##.
Hmm, if ##b\neq a## then it has at most two solutions right? Not an interval of solutions.

Delta2 said:
Hmm, if ##b\neq a## then it has at most two solutions right? Not an interval of solutions.
Yes. My point was really if we draw a graph of the function, then all is clear.

Delta2 and Mark44
It is necessary that the solutions lie in said interval. Conversely, given any ##x\in [0,2]## one readily verifies it satisfies the equality.

## What is modulus in mathematics?

Modulus, also known as absolute value, is a mathematical operation that returns the distance of a number from zero on a number line. It is always a positive value, regardless of the sign of the original number.

## How do you solve for x involving modulus?

To solve for x involving modulus, you must first isolate the modulus expression on one side of the equation. Then, you can solve for both the positive and negative values of the expression within the modulus brackets. The solutions will be the values of x that satisfy the original equation.

## Can an equation involving modulus have more than one solution?

Yes, an equation involving modulus can have more than one solution. This is because the modulus operation returns the distance from zero, so there can be both positive and negative values that satisfy the equation.

## What is the difference between modulus and absolute value?

Modulus and absolute value are essentially the same operation, but they are used in different contexts. Modulus is typically used in equations, while absolute value is used to represent the magnitude of a number.

## Can modulus be applied to complex numbers?

Yes, modulus can be applied to complex numbers. In this case, the modulus operation returns the distance of the complex number from the origin on a complex plane.

Replies
10
Views
1K
Replies
3
Views
1K
Replies
3
Views
958
Replies
7
Views
1K
Replies
14
Views
1K
Replies
11
Views
1K
Replies
2
Views
1K
Replies
16
Views
1K
Replies
6
Views
664
Replies
11
Views
1K