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Absolute Value Equations

Many students are uncomfortable with the concept of absolute value. They often think of it as some magic trick that makes everything positive. In a sense this is correct, but if that’s how you think about absolute value, then solving absolute value equations and inequalities will just be more complicated magic tricks. In other words, this view of absolute value is uninformative – it doesn’t give you any conceptual basis for understanding these problems.

So how should one think about absolute value?

Absolute value is really a geometric concept. You may have noticed that geometry problems only have positive answers – you can’t have negative distance, circumference, area, or volume… at least not in our universe.

The absolute value of a number is just it’s distance from zero on the number line. For example 5 and -5 have the same absolute value because they are both 5 units away from zero. We can use this information to solve relatively simple absolute value equations and inequalities.  Consider the following example:

\(

Because the absolute value of a number is just the distance between that number and zero on the number line, this equation has two solutions – [latex]x = 7\) and \(x=-7\). This can also be represented on a crude impersonation of a number line like so:

<——————–(-7)——————–(0)——————–(7)———————>

I know, I know – to easy! You’re drowning, and I’m describing the water… or something like that.

SOP for solving more complicated absolute value equations is to drop the absolute value and split the equation into two separate equations like so:

\(|x – 3| = 5\) becomes \(x – 3 = 5\) and \(x – 3 = -5\).

Now we do some algebra and we get [latexh]x = 8[/latex] and \(x = -2\)

There’s nothing wrong with that approach. ABSOLUTELY NOTHING WRONG! This approach will work for everything you see on the GRE or GMAT. However, I can’t resist presenting a more intuitive way to think about this. The approach above is simply a recipe for solving these equations. Fundamental understanding of concepts beats memorizing recipes (aka algorithms) hands down every time…

Absolute value can help us calculate the distance between arbitrary numbers on the number line. Given two values \(a\) and \(b\) I can find the distance between them by finding \(|a – b|\)

Now we can go back the our more complicated example \(|x – 3| = 5\), and think of it like this – the distance between \(x\) and \(3\) is \(5\). That is, \(x\) is a number that is five units away from three.

So the solutions must be \(3 + 5 = 8\) and \(3 – 5 = -2\).

I know, I know – still too easy, or, at least, not hard enough. You’re still drowning and my description of the water is getting more and more detailed. There are several wrinkles that need to be addressed.

(1) What about \(|x + 4| = 12\)

(2) And what about \(|2x + 4| = 12\)

(3) Oh yeah, what about \(|2x + 4| = |3x – 12|\)

Number (1) requires a little trick: \(x + 4 = x – (-4)\). So the equation can be written \(|x – (-4)| = 12\). So we’re looking for the numbers that are twelve units away from negative four – \(-4 + 12 = 8\) and \(-4 – 12 = -16\).

Number (2) is just another trick – you can factor positive numbers out of absolute value expressions: \(|2x + 4| = 2|x + 2| \). Dividing both sides by two gives us \(|x + 2| = 6\) and now we use trick number (1) – \(|x + 2| = |x – (-2)| = 6\). The distance between \(x\) and \(-2\) is \(6\). So \(x = -2 + 6 = 4\) or \(x = -2 – 6 = -8\).

Number (3) is a tough nut to crack using geometric thinking. So tough that I’m not going to put you through it. This is a case where you should use the recipe.

That might seem like a cop-out, but you should always know two or three, but probably not seven ways to solve a problem. If you’ve only got one tool in your tool box, your too limited to be the a flexible problem solver who will send the GMAT or GRE home cryin’ to its mama. Seven tools, on the other hand, is overkill for standardized tests, and will probably lead to choice paralysis.