Hello.
This lesson will discuss the solution of equations involving modulus. I’ll talk about linear equations mainly. But the methods you’ll learn will be applicable across any type of equation, be it linear, quadratic, trigonometric, or any other type.
Example 1 Solve the equation |x| = 3.
Solution I’ll illustrate two different methods, both following from the definitions that we discussed previously.
Method 1
The definition of modulus tells me that
\( |x| = \begin{cases}
x& \text{ if } x\ge 0 \\
-x& \text{ if } x < 0
\end{cases} \)
So when x ≥ 0, |x| can be replaced by x, and therefore I’ve to solve the equation x = 3.
And, when x < 0, |x| equals –x, and hence I’ve to solve –x = 3.
The first equation gives the solution straight away, i.e. x = 3. The second one gives x = –3.
We therefore get two solutions x = 3 and –3.
You can substitute these solutions back in the original equation to verify: |3| = 3 and |–3| = 3.
Method 2
We’ll now use the second definition. Recall that |x| represents the distance of x from 0 on the number line.
We can therefore rephrase this problem as “Find a number whose distance from the origin is 3”.
Well, that’s easy enough. Let me draw the number line.
On this number line, we can clearly see that there are two numbers, 3 and – 3, that are at a distance of 3 from the origin.
And we’re done. |x| = 3 gives x = 3 and –3. We generally write these two solutions as x = ± 3.
The second method is kind of a ‘visual’ approach to this problem. Though it seems easier, we might not be able to apply it in every problem.
The first method is more of a ‘theoretical’ or analytical approach. This method will become clearer as you go through more examples.
Let’s try another one.
Example 2 Find the value of y, where |y| = 5.
Solution As you may have guessed, the solutions to this equation are y = 5 and y = –5.
Why? Because, we’ve to find a number y whose distance from 0 (i.e. |y|) equals 5.
And seeing at the number line, the number can be 5 or –5.
There seems to be a pattern. Can we generalize?
Looking at an equation of the form |x| = a, we could directly say x = ±a.
For example, |x| = 7 implies x = ±7. And |x| = 1 gives x = 1 and – 1. Neat.
But wait! Before we put a stamp on it, consider the following two examples.
Example 3 Solve for x: |x| = 0.
Solution Can we say x = ±0? Well, we can. But 0 and –0 aren’t any different. And –0 looks kind of sloppy.
So, kicking away the ‘–0’, we’ll only keep x = 0 as the solution.
Therefore, |x| = 0 implies x = 0.
In other words, the only number that is a distance of 0 from 0 is 0 itself.
Coming back to |x| = a, what if ‘a’ is negative?
Example 4 Solve the equation |z| = –5.
Solution I know you’re tempted to write z = ± 5. But these values of z do not satisfy the given equation.
Substituting z = 5 in the LHS, we get |5| or 5. But RHS = –5.
Similarly, if we substitute z = –5 in the LHS, we get |–5| which equals 5 again, and is not equal to the RHS.
So how do we solve it then?
Recall again that the equation is asking you to find a number whose distance from the origin is –5.
Negative distance? That’s not possible. This means that there’s no possible value of z.
Or, the given equation has no solution.
Let’s try the analytical method.
First, we’ll assume z to be greater than or equal to zero, i.e., z ≥ 0.
Then the equation becomes z = –5. Or, we get the value of z as –5.
We get a negative value of z. But that’s against our assumption (i.e. z ≥ 0). Hence this is not a valid solution.
Not let’s assume z < 0. Since |z| = –z in this case, the equation becomes –z = –5, which gives z = 5.
Here, we obtained a positive value of z. This is again an invalid solution, as our assumption was z < 0.
We’ve exhausted all possible cases of z. Hence, the given equation has no solution.
Once again, the ‘visual’ approach seemed to be easier, but be sure that you understand the analytical method as well. Things will get complicated later on, when visualizing won’t be as easy as this.
Lesson Summary
We can now directly write the solution of the equation |x| = a as follows
- If a > 0, then |x| = a has the solution x = ±a.
- If a = 0, then |x| = 0 has the solution x = 0.
- If a < 0, then |x| = a has no solution.
I’ll cover a few more examples related to equations in the next lesson.