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Section 5.7 Radical Equations
Example 5.7.1 .
To solve the equation
\begin{equation*}
\sqrt{x} +4 =7
\end{equation*}
first isolate the radical term:
\begin{equation}
\sqrt{x} = 3\tag{5.7.1}
\end{equation}
Squaring both sides,
\begin{equation*}
x = 9
\end{equation*}
which we can confirm is the solution
\begin{equation*}
\sqrt{9}+4 = 3 + 4 = 7
\end{equation*}
The following example illustrates why we must be so careful here.
Example 5.7.3 . A radical equation.
Solve
\begin{equation*}
\sqrt{x-1} + x = 1.
\end{equation*}
We first isolate the radical term:
\begin{equation*}
\sqrt{x-1} = 1-x
\end{equation*}
Next, square both sides of the equation
\begin{equation*}
x-1 = (1-x)^2
\end{equation*}
Expanding,
\begin{equation*}
x-1 = 1-2x + x^2
\end{equation*}
Combine like terms
\begin{equation*}
x^2 -3x +2= 0
\end{equation*}
which is quadratic and can be factored as
\begin{equation*}
(x-2)(x-1) = 0
\end{equation*}
So the solutions to this quadratic equation are \(x = 2\) and \(x = 1\text{.}\)
But be careful! We need to check each one in the original equation.
\begin{equation*}
\sqrt{x-1} + x = 1
\end{equation*}
For instance, when \(x=2\text{,}\)
\begin{equation*}
\sqrt{2-1} + 2 = \sqrt{1}+2 = 1+2 = 3 \neq 1
\end{equation*}
so this is not a solution after all! Checking \(x=1\text{,}\)
\begin{equation*}
\sqrt{1-1} + 1 = \sqrt{0} + 1 = 0 + 1 = 1.
\end{equation*}
Thus, the only solution to our equation is \(x=1\text{.}\)
