Radical Equations

Task

1. Solve the following two equations by isolating the radical on one side and squaring both sides:

   1. $\sqrt{2x+1}-5=-2$

   2. $\sqrt{2x+1}+5=2$

   Be sure to check your solutions.

2. If we raise both sides of an equation a power, we sometimes obtain an equation which has more solutions than the original one. (Sometimes the extra solutions are called extraneous solutions.) Which of the following equations result in extraneous solutions when you raise both sides to the indicated power ? Explain.

   1. $\sqrt{x}=5$, square both sides

   2. $\sqrt{x}=-5$, square both sides

   3. $\sqrt[3]{x}=5$, cube both sides

   4. $\sqrt[3]{x}=-5$, cube both sides

3. Create a square root equation similar to the one in part (a) that has an extraneous solution. Show the algebraic steps you would follow to look for a solution, and indicate where the extraneous solution arises.

IM Commentary

In order to engage this task meaningfully, students must be aware of the convention that $\sqrt{a}$ for a positive number $a$ refers to the positive square root of $a$. The purpose of the task is to show students a situation where squaring both sides of an equation can result in an equation with more solutions than the original one. The reason for this is that it is possible to have two unequal numbers whose squares are equal (for example, $3$, and $-3$). So if the squares of each side of an equation are equal for some value of the variable, that does not mean the original sides were equal. Cube roots and cubing do not have the same problem, because if the cubes of two numbers are equal then the original numbers are equal.

Solution

1. 1.   $$ \begin{align} \sqrt{2x+1} - 5 &= -2 \\ \left( \sqrt{2x+1}\right)^2 &= (3)^{2} \\ 2x+1 &= 9 \\ 2x &= 8 \\ x &= 4 \end{align} $$

      Checking:

        $$ \begin{align} \sqrt{2 \cdot 4+1} - 5 &= -2 \\ \sqrt{9} - 5 &= -2 \\ 3 - 5 &= -2 \\ -2 &= -2 \end{align} $$ So $x=4$ is the solution to the equation.

   2.   $$ \begin{align} \sqrt{2x+1} + 5 &= 2 \\ \left( \sqrt{2x+1}\right)^2 &= (-3)^{2} \\ 2x+1 &= 9 \\ 2x &= 8 \\ x &= 4 \end{align} $$

      Checking:

        $$ \begin{align} \sqrt{2 \cdot 4+1} + 5 &= 2 \\ \sqrt{9} + 5 &= 2 \\ 3 + 5 &= 2 \\ 8 &\neq 2 \end{align} $$ So this equation has no solution.

2. The only one of the equations that produces an extraneous solution is

   $$\sqrt{x}=-5$$

   The square root symbol (like all even roots) is defined to be the positive square root, so a positive root can never be equal to a negative number. Squaring both sides of the equation will make that discrepancy disappear; the square of a positive number is positive but so is the square of a negative number, so we’ll end up with a solution to the new equation even though there was no solution to the original equation.

   This isn’t the case with odd roots - a cube root of a positive number is positive, and a cube root of a negative number is negative. When we cube both sides of the last equation, the negative remains, and we end up with a true solution to the equation.

3. If we start with a square root set equal to a negative number, we know that squaring both sides will get us into trouble. Something like $$\sqrt{2x+1}=-5$$ will work. If we can add a positive number to both sides, it will be less obvious. Perhaps: $$\sqrt{2x+1}+7=2.$$ Solving this equation: $$ \begin{align} \sqrt{2x+1}+7&=2 \\ \left(\sqrt{2x+1}\right)^2 &=(-5)^2 *\\ 2x + 1&= 25 \\ 2x &= 24 \\ x &= 12 \end{align} $$

   * - This is where the extraneous solution comes in. The square root can’t be negative, but by squaring both sides, we’re losing that information.