Problem 26 - Introductory Algebra for College Students

Question

In solving \(\sqrt{2 x-1}+2=x,\) why is it a good idea to isolate the radical term? What if we don't do this and simply square each side? Describe what happens.

Step-by-Step Solution

Answer

Isolating the radical term is crucial to ensure that we do not introduce any extraneous roots into the solution. In the given exercise, it was found that not isolating the radical term led to an extraneous root \(x = 5\), which did not validate when checked in the original equation. The valid root is \(x = 1\).

Step 1: Isolate the Radical Term

In order to properly simplify the equation without creating extra roots, it is best to isolate the radical term first. To do this, subtract 2 from both sides of the equation, ending up with \(\sqrt{2 x-1}=x-2\).

Step 2: Square Both Sides

To get rid of the square root, square both sides of the equation. This yields \(2 x - 1 = (x - 2)^2.\) After expanding the right hand side, we obtain \(2 x - 1 = x^2 - 4x + 4 \).

Step 3: Simplify the Quadratic Equation

Rearrange the terms to form a quadratic equation, resulting in \(x^2 - 6x + 5 = 0\).

Step 4: Solve the Quadratic Equation

Solve the quadratic equation using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In this context, \(a = 1\), \(b = -6\), and \(c = 5\). So, the solutions are \(x = 1\) or \(x = 5\).

Step 5: Validate Solutions

Check whether both solutions are valid by substituting them back into the original equation. In this case, plugging \(x = 1\) validates, but substituting \(x = 5\) does not. The reason behind it is that \(5\) is an extraneous root that occurred because of squaring both sides in step 2.