Lesson 4: Linear Equations in Disguise

Property

Solving radical equations containing an even index by raising both sides to the power of the index may introduce an algebraic solution that would not be a solution to the original radical equation. This is called an extraneous solution. Always check your answer in the original equation.

Examples

• Solve $\sqrt{r+1} = r-1$. Squaring gives $r+1 = r^2-2r+1$, which simplifies to $r^2-3r=0$, or $r(r-3)=0$. The algebraic solutions are $r=0$ and $r=3$. Checking $r=0$ gives $\sqrt{1}=-1$ (False). Checking $r=3$ gives $\sqrt{4}=2$ (True). Thus, $r=0$ is an extraneous solution.

• Solve $\sqrt{m+9} - m + 3 = 0$. Isolate the radical: $\sqrt{m+9} = m-3$. Squaring gives $m+9 = m^2-6m+9$, so $m^2-7m=0$. The solutions are $m=0$ and $m=7$. Only $m=7$ is a valid solution.

Property

When multiplying an equation by an expression containing a variable, a false solution may be introduced. Such a solution is called an extraneous solution. To check for extraneous solutions, substitute the possible solution into the original equation. If it causes any denominator in the original equation to equal zero, that solution is extraneous and must be discarded.

Examples

• Solve $5 + \frac{2}{x-4} = \frac{x-2}{x-4}$. Multiplying by the LCD $x-4$ gives
  $$5(x-4) + 2 = x-2$$
  , so
  $$5x - 20 + 2 = x-2$$
  , which simplifies to
  $$4x = 16$$
  and
  $$x=4$$
  . Since $x=4$ makes the original denominator zero, it is an extraneous solution. There is no solution.
• Solve $\frac{y}{y+2} - 1 = \frac{-2}{y+2}$. The LCD is $y+2$. This gives
  $$y - 1(y+2) = -2$$
  , so
  $$y - y - 2 = -2$$
  , which is
  $$-2=-2$$
  . The proposed solution is $y=-2$, which is extraneous. No solution.
• Solve $\frac{3a}{a-6} - 4 = \frac{18}{a-6}$. The LCD is $a-6$. This gives
  $$3a - 4(a-6) = 18$$
  , so
  $$3a - 4a + 24 = 18$$
  , which simplifies to
  $$-a = -6$$
  and
  $$a=6$$
  . This solution is extraneous.

Explanation

An extraneous solution is a tricky fake-out! It appears to be a valid answer, but it's invalid in the original context because it makes you divide by zero. Always plug your solution back in to check the original denominators.