Example 2
Completion requirements
| Example 2 |
Solve \(\sqrt {2x} = 8\).
Step 1: Identify any restrictions on the variable.
Because the index is \(2\) (even), the radicand must be greater than or equal to zero.
\(\begin{align}
2x &\ge 0 \\
x &\ge 0, x \in \rm R \\
\end{align}\)
Step 2: Isolate the radical.
In this example, this is already done.
Step 3: Raise both sides of the equation to the power of the index in order to remove the radical.
In this example, the index is two, so raise both sides of the equation to the exponent of two.
\(\begin{align}
\sqrt {2x} &= 8 \\
\left( {\sqrt {2x} } \right)^2 &= 8^2 \\
2x &= 64 \\
\end{align}\)
Step 4: Solve for the variable.
\(\begin{align}
2x &= 64 \\
x &= 32 \\
\end{align}\)
Check that this value is within the variable’s restrictions.
Because \(32\) is greater than zero, the solution is within the restriction.
Step 5: Verify the solution.
The solution is \(x = 32\).
Because the index is \(2\) (even), the radicand must be greater than or equal to zero.
\(\begin{align}
2x &\ge 0 \\
x &\ge 0, x \in \rm R \\
\end{align}\)
Step 2: Isolate the radical.
In this example, this is already done.
Step 3: Raise both sides of the equation to the power of the index in order to remove the radical.
In this example, the index is two, so raise both sides of the equation to the exponent of two.
\(\begin{align}
\sqrt {2x} &= 8 \\
\left( {\sqrt {2x} } \right)^2 &= 8^2 \\
2x &= 64 \\
\end{align}\)
Step 4: Solve for the variable.
\(\begin{align}
2x &= 64 \\
x &= 32 \\
\end{align}\)
Check that this value is within the variable’s restrictions.
Because \(32\) is greater than zero, the solution is within the restriction.
Step 5: Verify the solution.
| Left Side | Right Side |
|---|---|
| \(\begin{array}{r} \sqrt {2x} \\ \sqrt {2\left( {32} \right)} \\ \sqrt {64} \\ 8 \\ \end{array}\) | \(8\) |
| LS = RS | |
The solution is \(x = 32\).