Integrated Math 3 / Unit 4: Radical Functions and Equations
4.5 Extraneous Solutions
Notes
Squaring both sides can create answers that do not satisfy the original equation. These are extraneous solutions.
- Example: $\displaystyle \sqrt{x + 6} = x$ gives $x + 6 = x^2$, so $(x - 3)(x + 2) = 0$. Check $x = 3$: $\displaystyle \sqrt{9} = 3$. Check $x = -2$: $\displaystyle \sqrt{4} = 2 \neq -2$. So only $x = 3$ works.
- A principal square root is never negative, so if the other side is negative the answer is extraneous.
- Always substitute every candidate into the original equation.
Practice
Solve $\displaystyle \sqrt{x + 6} = x$.
The candidates are 3 and $-2$. For $x = -2$ the left side is $2$ and the right side is $-2$, so it is extraneous.
Why can squaring both sides of an equation produce extraneous solutions?
For example, $x = -2$ and $x = 2$ both satisfy $x^2 = 4$, but only one may satisfy the original equation.
Solve $\displaystyle \sqrt{3 - x} = x - 3$.
Squaring gives $3 - x = x^2 - 6x + 9$, so $x^2 - 5x + 6 = 0$ and $x = 2$ or $3$. For $x = 2$: $\displaystyle \sqrt{1} = 1 \neq -1$. For $x = 3$: $0 = 0$.
Ready for more? Take the Unit 4 test.