Example 5
Completion requirements
| Example 5 |
Solve \(\sqrt {x - 5} + 5 = 4\).
Step 1: Identify any restrictions on the variable.
Only the radicand is restricted; therefore \(x - 5 \ge 0\), or \(x \ge 5, x \in \rm R\).
Step 2: Isolate the radical.
\(\begin{align}
\sqrt {x - 5} + 5 &= 4 \\
\sqrt {x - 5} &= -1 \\
\end{align}\)
Step 3: Raise both sides of the equation to the power of the index in order to remove the radical, and then solve for the variable.
\(\begin{align}
\sqrt {x - 5} &= -1 \\
\left( {\sqrt {x - 5} } \right)^2 &= \left( { - 1} \right)^2 \\
x - 5 &= 1 \\
x &= 6 \\
\end{align}\)
The solution is within the variable’s restrictions.
Step 4: Verify the solution.
Because the left side does not equal the right side, \(x = 6\) does not satisfy the equation and is considered an extraneous root. There is no solution to this equation.
Only the radicand is restricted; therefore \(x - 5 \ge 0\), or \(x \ge 5, x \in \rm R\).
Step 2: Isolate the radical.
\(\begin{align}
\sqrt {x - 5} + 5 &= 4 \\
\sqrt {x - 5} &= -1 \\
\end{align}\)
Step 3: Raise both sides of the equation to the power of the index in order to remove the radical, and then solve for the variable.
\(\begin{align}
\sqrt {x - 5} &= -1 \\
\left( {\sqrt {x - 5} } \right)^2 &= \left( { - 1} \right)^2 \\
x - 5 &= 1 \\
x &= 6 \\
\end{align}\)
The solution is within the variable’s restrictions.
Step 4: Verify the solution.
| Left Side | Right Side |
|---|---|
| \( \begin{array}{r} \sqrt {x - 5} + 5 \\ \sqrt {6 - 5} + 5 \\ \sqrt 1 + 5 \\ 1 + 5 \\ 6 \\ \end{array}\) | \(4\) |
| LS \(\ne\) RS \(\hspace{30pt}\) | |
Because the left side does not equal the right side, \(x = 6\) does not satisfy the equation and is considered an extraneous root. There is no solution to this equation.