# Extraneous Solutions

## Key Questions

• A lot of times in algebra, especially when you deal with radical functions, you will end up with what you call extraneous solutions. These are solutions to an equation that you will get as a result of your algebra, but are still not correct. It's not that your process is wrong; it's just that this solution does not fit back into the equation (math is very complicating sometimes).

To find whether your solutions are extraneous or not, you need to plug each of them back in to your given equation and see if they work. It's a very annoying process sometimes, but if employed properly can save you much grief on tests or quizzes.

There are two ways you can do this: by hand, and using your graphing calculator. I will go over both, in case you don't have and/or are not allowed to use one on tests/quizzes. In any case, it is helpful to know both so you can get a better feel of how it works, and just to be that much better at math :)

By Hand:

Consider the equation $\sqrt{x + 4} = x - 2$

First, let's solve it using usual algebra:

${\left(\sqrt{x + 4}\right)}^{2} = {\left(x - 2\right)}^{2}$ (square both sides)

$x + 4 = {x}^{2} - 4 x + 4$ (simplify)

${x}^{2} - 5 x = 0$ (subtract $x$ and 4 from both sides)

$x \left(x - 5\right) = 0$. hence $x = 0 \mathmr{and} x = 5$ (zero product property)

Now we have our two solutions; 0 and 5. Now let's plug each of them back into our original equation and see if they work:

5:

$\sqrt{5 + 4} = 5 - 2$
$\sqrt{9} = 3$
$3 = 3$

Therefore, 5 is a verified solution.

0:

$\sqrt{0 + 4} = 0 - 2$
$\sqrt{4} = - 2$
2 ≠ -2

Therefore, 0 is not a solution.

Hence, we would classify 0 as an extraneous solution to this given equation.

By Calculator:

• Set the equation to equal zero
(this ends up being $\sqrt{x + 4} - x + 2 = 0$)
• Plug this into the $y =$ button on your TI-83/84 calculator
• Find the value of each of your solutions (go to 2nd->Calc->Value and enter your solution for $x$)
• You should get zero as an answer for each of them. If you don't, that solution is extraneous.

This is often a much easier way to do it, but as I said above, it is important that you know how to do it by hand as well, as often teachers may ask you to show work, and you may not always have a calculator to help you out.

Hope that helped :)

*equation from http://hotmath.com/*

• In general, extraneous solutions arise when we perform non-invertible operations on both sides of an equation. (That is, they sometimes arise, but not always.)

Non-invertible operations include: raising to an even power (odd powers are invertible), multiplying by zero, and combining sums and differences of logarithms.

Example :
The equations: $x + 2 = 9$ and $x = 7$, have exactly the same set of solutions. Namely: $\left\{7\right\}$.

Square both sides of $x = 7$ to get the new equation: ${x}^{2} = 49$. The solution set of this new equation is; $\left\{- 7 , 7\right\}$. The $- 7$ is an extraneous solution introduced by squaring the two expressions

Square both sides of $x + 2 = 9$ to get the new equation: ${x}^{2} + 4 x + 4 = 81$. Solve the new equation:
${x}^{2} + 4 x - 77 = 0$ so $\left(x - 7\right) \left(x + 11\right) = 0$ whose solution set is $\left\{7 , - 11\right\}$. The $- 11$ does not solve the original equation.

• Example 1 : Raising to an even power
Solve $x = \sqrt[4]{5 {x}^{2} - 4}$.
Raising both sides to the ${4}^{t h}$ gives ${x}^{4} = 5 {x}^{2} - 4$.
This requires, ${x}^{4} - 5 {x}^{2} + 4 = 0$.
Factoring gives $\left({x}^{2} - 1\right) \left({x}^{2} - 4\right) = 0$.
So we need $\left(x + 1\right) \left(x - 1\right) \left(x + 2\right) \left(x - 2\right) = 0$.

The solution set of the last equation is $\left\{- 1 , 1 , - 2 , 2\right\}$. Checking these reveals that $- 1$ and $- 2$ are not solutions to the original equation. Recall that $\sqrt[4]{x}$ means the non-negative 4th root.)

Example 2 Multiplying by zero
If you solve $\frac{x + 3}{x} = \frac{5}{x}$ by cross multiplying,
you'll get ${x}^{2} + 3 x = 5 x$
which lead to ${x}^{2} - 2 x = 0$
.
It looks like the solution set is $\left\{0 , 2\right\}$.
Both are solutions to the second and third equations, but $0$ is not a solution to the original equation.

Example 3 : Combining sums of logarithms.
Solve: $\log x + \log \left(x + 2\right) = \log 15$
Combine the logs on the left to get $\log \left(x \left(x + 2\right)\right) = \log 15$
This leads to $x \left(x + 2\right) = 15$ which has 2 solutions: $\left\{3 , - 5\right\}$. The $- 5$ is not a solution to the original equation because $\log x$ has domain $x > 0$ (Interval: $\left(0 , \infty\right)$)

• In general, extraneous solutions arise when we perform non-invertible operations on both sides of an equation. (That is, they sometimes arise, but not always.)

Squaring (or raising to any other even power) is a non-invertible operation. Solving equations involving square roots involves squaring both sides of an equation.

Example 1 : To show the idea:
The equations: $x - 1 = 4$ and $x = 5$, have exactly the same set of solutions. Namely: $\left\{5\right\}$.

Square both sides of $x = 5$ to get the new equation: ${x}^{2} = 25$. The solution set of this new equation is; $\left\{- 5 , 5\right\}$. The $- 5$ is an extraneous solution introduced by squaring the two expressions

Square both sides of $x - 1 = 4$ to get ${x}^{2} - 2 x + 1 = 16$
which is equivalent to ${x}^{2} - 2 x - 15 = 0$.
and, rewriting the left, $\left(x + 3\right) \left(x - 5\right) = 0$.
So the solution set is $\left\{- 3 , 5\right\}$.
This time, it is $- 3$, that is the extra solution.

Example2 : Extraneous solution.
Solve $x = 2 + \sqrt{x + 18}$
Subtracting $2$ from both sides: $x - 2 = \sqrt{x + 18}$

Squaring (!) gives ${x}^{2} - 4 x + 4 = x + 18$
This requires, ${x}^{2} - 5 x - 14 = 0$.
Factoring to get $\left(x - 7\right) \left(x + 2\right) = 0$

finds the solution set to be $\left\{7 , - 2\right\}$.
Checking these reveals that $- 2$ is not a solution to the original equation. (It is a solution to the 3rd equation -- the squared equation.)

Example 3 : No extraneous solution.
Solve $\sqrt{{x}^{2} + 9} = x + 3$
Squaring (!) gives, ${x}^{2} + 9 = {\left(x + 3\right)}^{2} = {x}^{2} + 6 x + 9$
Which leads to $0 = 6 x$ which has only one solution, $0$ which works in the original equation.