4. Equations Involving Modulus (Part 2)

In this lesson, I’ll continue with equations involving modulus. We’ll build upon the knowledge gained in the previous lesson.

Example 1 Solve for x: |x – 4| = 5.

Solution If you figured out the next step to be x – 4 = ± 5, awesome!

Don’t worry if you couldn’t. I was going to explain it anyway.

We’re already familiar with equation of the form |x| = 3. The solution to this equation is x = ± 3.

In the case of the given equation, we can easily convert it to this familiar form.

That is, we’ll substitute x – 4 = y, and our equation will look like |y| = 5. And now we can solve it easily.

We get y = ± 5, which gives us x – 4 = ± 5 or x = ± 5 + 4.

Finally, we get the values of x as 9 and –1. To verify, substitute these values in the original equation if you wish to.

Example 2 Find the value of x if |x + 3| = 0.

Solution Once again, we can substitute x + 3 as y (or z, if you like) and get |y| = 0. And from the previous lesson, we know that this has the solution y = 0, which gives x + 3 = 0 or x = –3.

Example 3 Find the value of x if |x – 6| = –2.

Solution I hope you got the idea by now. The given equation has no solution.

We don’t even need to make any substitution. |x| is always non-negative. In fact, the modulus of any expression is always non-negative.

But the problem is asking for a value of x for which |x – 6| is negative, which isn’t possible.

Hence, we’ll straight away conclude that there’s no solution to the equation.

Example 4 Find the value of x for which |x – 5| = 2.

Solution Why the same thing again? Because I want to use another method to approach the problem.

Recall that |x| denotes the distance of the number x from the origin. Well, that means |x – 5| denotes the distance of x – 5 from the origin.

Let’s visualise this on the number line. We’ll take any number x on the line.

Then, x – 5 is the number that lies 5 units towards the left of x.

And, |x – 5| is nothing but the distance of this x – 5 from the origin.

Hmm. That’s uninteresting. But if you observe closely, this |x – 5| is equal to another distance.

That is, |x – 5| is the distance between x and 5. Pretty neat!

Do you see why? Think of shifting both x and 5 by five units towards the left. Then, x reaches x – 5 and 5 reaches 0.

And since we shifted both of them by the same distance (doesn’t matter whether left or right), the distance between them will remain the same (i.e., before and after shifting).

Let’s come back to our equation |x – 5| = 2. It can now talk.

It says, “Find the number x, which is at a distance of 2 units from the number 5 on the number line.”

And I reply, “There are two such values of x: 5 + 2 and 5 – 2.” Or, 7 and 3, as you can see below.

Let’s try another one.

Example 5 Find the value of x for which |x + 7| = 3.

Solution Remember that |x – a| is the distance between x and a on the number line. Here, we have ‘+’ sign instead of a ‘–’. That’s not a problem. We’ll simply rewrite the expression on the left as |x –(–7)|.

Now this equation says: Find the number x, which is at a distance of 3 units from –7.

There are two values again: –7 + 3 and –7 – 3, or –4 and –10. Have a look.

Once again, this visual method wins.

But as I mentioned earlier, this might not always work, or atleast things might not be as easy to visualise.

Say, for example, |x2 – 5| = 4. It isn’t very obvious where to place x2 on the number line, given where x lies. In such cases, we’ll use the analytical method, as we did in the first three examples. Let me show you how.

Example 6 Solve for x: |x2 – 5| = 4.

Solution As already mentioned, I’ll use the method which was used in the first example.

I’m directly going to write x2 – 5 = ±4, which gives x2 = ±4 + 5.

From here, we get x2 = 9 or 1. Finally, we get the values of x as 1, –1, 3 and –3.

Let’s do a couple more.

Example 7 Solve for x: |x2 – 4| = 5.

Solution This gives x2 – 4 = ± 5, which leads to x2 = ± 5 + 4 or 9 and –1.

Now, x2 = –1 doesn’t have any real solutions, while x2 = 9 gives x = ± 3.

Example 8 Solve for x: |x2 – 5x| = 6.

Solution This gives x2 – 5x ± 6 = 0. I’ll solve the two quadratic equations separately.

The first one, x2 – 5x + 6 = 0, becomes (x – 2)(x – 3) = 0, which has the roots 2 and 3.

The second one, x2 – 5x – 6 = 0, becomes (x – 6)(x + 1) = 0, which gives x = 6 and –1.

Therefore, we get four values of x satisfying the given equation: –1, 2, 3 and 6.

Okay, enough. Last one. This one is a bit more interesting.

Example 9 Solve for x: |x2 + 5| = 4.

Solution We get x2 + 5 = ± 4, which gives x2 = ±4 – 5.

We finally get the equations x2 = –9 and x2 = –1, both of which don’t have any real solutions.

Here’s the interesting part. We could have concluded that just by looking at the equation!

The term x2 will always be non-negative (for real x). That is, x2 ≥ 0 for all x ϵ R.

What about x2 + 5 then?

Well, if x2 is always greater than or equal to zero (i.e. x2 ≥ 0), then x + 5 will be always greater than or equal to 5 (i.e. x2 + 5 ≥ 5).

And what about |x2 + 5|? Let me simplify a bit by substituting x2 + 5 as y.

If y is a number that’s greater than or equal to 5, then its distance from the origin, or |y|, will also be greater than or equal to 5.

That means, |x2 + 5| will always be greater than or equal to 5, whatever be the value of x.

And the equation is trying to equate it to 4, which is never possible. Hence, the given equation has no real solutions.

Pay attention to ‘real solutions’ there. Had x been a complex number, we may have obtained some solutions. But that’s not something we’ll discuss here. Assume throughout the series that x (or whichever variable I use) is a real number.

By the way, I gave you a sneak peek into inequalities involving modulus, something which I’ll be covering after equations.

The next lesson will deal with some complicated equations involving modulus. See you there.

P.S. I’m going to discuss a third method of solving such equations sometime later, when I discuss graphs. That’ll turn out to be the most interesting one.