Could someone please explain how to identify the extraneous solution and what it means?

Question

jdoe0001 · Answer

got any specifics?

anonymous · Answer

an extraneous solution is a solution that either is a negative number when you're working with distance (ex. you can't have -5 feet), or when you plug it back into the equation you get something like 1=1

mathstudent55 · Answer

An extraneous solution is a false solution. Sometimes, when you solve and equation involving a square root, the method of solving requires squaring both sides. Then you solve the new equation. Squaring both sides can introduce solutions that are true solutions to the new squared equation, but were not solutions to the original equation. These false solutions are called extraneous. This is why after solving an equation by squaring both sides, all solutions must be checked in the original equation.

mathstudent55 · Answer

Example:

Solve the radical equation: $\sqrt{x+4} = x + 2$

$\sqrt{x+4} = x + 2$

Square both sides:

$(\sqrt{x+4})^2 = (x + 2)^2$

$x + 4= x^2 +4x + 4$

$0 = x^2  + 3x $

$x^2 + 3x = 0$

$x(x + 3) = 0$

$x = 0$ or $x = -3$

Now let's check the solutions in the original equation.

1) Check x = 0

$\sqrt{x+4} = x + 2$

$\sqrt{0+4} = 0 + 2$

$\sqrt{4} = 2$

$2 = 2$ This is a true statement, so x = 0 is a solution to the original equation.

2) Check x = -3

$\sqrt{x+4} = x + 2$

$\sqrt{-3+4} = -3 + 2$

$\sqrt{1} = -1$

$1 = -1$ This is a false statement, so x = -3 is not a solution to the original equation.
x = -3 is an extraneous solution that was introduced by squaring both sides of the original equation.

In fact, if you check x = 0 and x = -3 in the squared equation,

$(\sqrt{x+4})^2 = (x + 2)^2$

you will see that both solutions work. We did not make a mistake in solving the original equation. It is simply that the method of squaring both sides sometimes introduces extraneous solutions.