Example 6

Step 1: Identify all restrictions on the variable.


Using a number line, the overall restriction on the variable can be determined.


This indicates that the restriction on the variable is \(x \ge 1, x \in \rm R\).

Step 2: Isolate one of the radical expressions.

In this example, one of the radicals, \(2\sqrt {x - 1}\), is already isolated. You can leave the two with the radical as well.

\(2\sqrt {x - 1} = \sqrt x - 1\)

Step 3: Square both sides of the equation to remove the first radical, then simplify until the second radical is isolated.

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

Step 4: Square both sides to remove the second radical, then simplify.

\(\begin{align}
 3x - 5 &= -2\sqrt x  \\ 
 \left( {3x - 5} \right)^2 &= \left( { - 2\sqrt x } \right)^2  \\ 
 9x^2 - 30x + 25 &= 4x \\ 
 9x^2 - 34x + 25 &= 0 \\ 
 \end{align}\)

Step 5: To solve the resulting quadratic equation, choose one of the methods learned in Unit 2.

Using the quadratic formula:

\(\begin{align}
 x &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\ 
 x &= \frac{{34 \pm \sqrt {\left( {-34} \right)^2 - 4\left( 9 \right)\left( {25} \right)} }}{{2\left( 9 \right)}} \\ 
 x &= \frac{{34 \pm \sqrt {1156 - 900} }}{{18}} \\ 
 x &= \frac{{34 \pm \sqrt {256} }}{{18}} \\ 
 x &= \frac{{34 \pm 16}}{{18}} \\ 
 x &= \frac{{34 + 16}}{{18}}{\rm { and }}x = \frac{{34 - 16}}{{18}} \\ 
 x &= \frac{{50}}{{18}}{\rm { and }}x = \frac{{18}}{{18}} \\ 
 x &= \frac{{25}}{9}{\rm { and }}x = 1 \\ 
 \end{align}\)

The solution is within the variable’s restrictions.

Step 6: Verify.