Exercise 12 Page 579

We are told that $x=2$ and $x=8$ are the solutions that were obtained when solving the following radical equation. We want to show that one of these solutions must be extraneous. Remember that square roots are only defined for non-negative numbers. Therefore, the square root expression $\sqrt{ax+b}$ will only produce non-negative values. Since $\sqrt{ax+b}$ is equal to $x-5,$ we know that $x-5$ can also only be equal to non-negative values. With this mind, let's solve for $x$ in the non-radical inequality. Great! This solution set means that the real solutions of the given equation must be greater than or equal to $5.$ Now, let's think about how the found solutions compare to $5.$ As we can see, $5$ is between the obtained solutions. Since we know that any real solutions to the equation must be greater than or equal to $5,$ and we know that $2$ is less than $5,$ we can conclude that $2$ must be an extraneous solution. Let's substitute $x=2$ back into the original equation and check our conclusion! A square root expression cannot be equal to a negative number so we know that our simplification lead to a contradiction. Therefore, $x=2$ does not satisfy the given radical equation and is definitively extraneous.