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## Ask Professor Puzzler

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Esther from Georgia wants to know "Why do extraneous solutions happen in math problems?"

Extraneous solutions are solutions which are obtained algebraically, but even though they are found algebraically, they don't work in the context of the problem. Here are some of the reasons a problem might have extraneous solutions. For each reason, I'll give an example.

## Real World Restrictions

Sometimes a solution that we find algebraically is extraneous because it doesn't fit real world restrictions on the solution. For example, consider the following problem:

A rectangle's length is 2 more than its width. Its area is 48 square feet. What is the rectangle's width?

To solve this, we say, let w equal the width. Then we have: w(w + 2) = 48

This is a quadratic equation: w2 + 2w - 48 = 0, which leads to w = 6 or w = -8. However, the real world restriction on the solution is that w is the width of a rectangle, and rectangles can't have negative side lengths. Therefore, -8 is an extraneous solution.

## Problem Restrictions

In other cases, the problem itself might set restrictions on the solution.

Find all two digit numbers such that when the number is squared, and ten times the number is subtracted, the result is 11.

This problem leads to the quadratic: x2 - 10x - 11 = 0

(x - 11)(x + 1) = 0

x = 11 or x = -1.

But the problem itself sets the restriction that x must be a two digit number, so x = -1 is extraneous.

## Division By Zero

Sometimes a problem has fractions, and a number that works out algebraically actually causes a division by zero, which is illegal, immoral, and socially unacceptable. As a problem writer for math competitions, I like to deliberately create problems in which this happens, because the very best students check their solutions and realize they've got an extraneous solution. Here's a very simple example:

(x + 1)/(x - 1) = 2/(x - 1)

An easy way to solve this is to multiply both sides by (x - 1), which gives x + 1 = 2, or x = 1. But the process of multiplying by (x - 1) has made a fundamental change to the problem; we no longer have a denominator. If we go back to the original problem and try to plug x = 1 into it, we find that we have a division by zero, and therefore x = 1 doesn't actually work; it's extraneous.

## Principle Roots

Each positive number has two real square roots. For example, the square roots of 4 are 2 and -2. However, when we write radical notation, we are, be definition, referring to the principle square root, which is the positive value. Thus, a solution may be extraneous because it results from using a negative square root instead of the principle square root. Here's an example:

√(x)  = x - 6

To solve this, we square both sides:

x = x2 - 12x + 36

Rearrange, simplify, factor, solve:

x2 - 13x + 36 = 0

(x - 9)(x - 4) = 0

x = 9 or x = 4

Now, plug 9 into the original equation, and you'll see that it does work. But what happens when you plug in 4? You get a false equation! But why does that happen? Because on the left-hand side, 4 has two square roots, and one of them (-2) does work. So it's our own choice to define √(x) as the positive square root of x that results in the extraneous solution.

Are there other causes for extraneous solutions? Probably, but that should serve as a simple crash course on the common ones you'll see in an algebra class.

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