Question

For what values of x,|x|+ x = 0? Explain.

Short answer

When x is positive, the solution is non zero. When x is any negative number, the solution set is any non-positive number and zero inclusive.

Step-by-step solution

– What is Modulus Function? 

The modulusfunction, which is also called the absolute value of a function, gives the magnitude or absolute value of a number irrespective of the number being positive or negative.

It always gives a non-negative value of any number or variable. Modulus function is denoted as y = |x| or f(x) = |x|, where f: R → R and x∈R.

|x| is the modulus of x, where x is a real number.

If x is non-negative then f(x) will be of the same value x.

If x is negative, then f(x) will be the magnitude of x, that is, f(x) = – x if x is negative

Modulus Function Formula: 

Modulus Function Formula

The value of the modulus function is always non-negative. If f(x) is a modulus function, then we have:

If x ≥ 0, then f(x) = x

If x = 0, then f(x) = 0

If x < 0, then f(x) = – x

This means if the value of x is greater than or equal to 0, then the modulus function takes the actual value, but if x is less than 0 then the function takes minus of the actual value 'x'.

For what values of x, |x| + x = 0? 

Case 1: is positive. The absolute value of x will be positive.

We can write: $\left|x\right|=x$ $\begin{array}{l}\left|x\right|+x=0\\ x+x=0\\ 2x=0\\ x=0÷2\end{array}$

We get x = 0

Case 2: If x is negative. The absolute value of will be positive.

We can write:$|-x|=x$$\begin{array}{l}|-x|+(-x)=0\\ x-x=0\\ 0=0\end{array}$

The solution for case 1 is and for case 2 is any negative number.

If x is less than 0 then the function takes minus of the actual value 'x'.

When x is negative or less than zero,|– x| + (– x) = x – x = 0

Example: When x = –1, then we can write |x | as |–1 | as and x = – 1

Then we can write: |– x| + (– x) = |– 1| + (– 1) = 1 + (– 1) = 1 – 1 = 0

Any negative number can satisfy the equation|x| + x = 0

The solution set is any non-positive number and zero inclusive.