# What are some examples of extraneous solutions to equations?

Mar 8, 2015

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: $\left(0 , \infty\right)$)