Problem 116 - College Algebra Essentials

Question

In solving \(\sqrt{3 x+4}-\sqrt{2 x+4}=2,\) why is it a good idea to isolate a radical term? What if we don't do this and simply square each side? Describe what happens.

Step-by-Step Solution

Answer

Isolating a radical before squaring sides simplifies the equation and makes it less prone to errors. If a radical isn't isolated before squaring, you may end up with a more complicated equation difficult to solve and could potentially produce invalid solutions.

Step 1: Isolate a radical

Rearrange the equation to isolate a root on one side - in this case \(\sqrt{3x+4}\). The equation becomes \(\sqrt{3x+4} = \sqrt{2x+4} + 2\).

Step 2: Square both sides

When both sides of the equation are squared, the root is eliminated on both sides. After squaring, the equation becomes \((\sqrt{3x+4})^2 = (\sqrt{2x+4}+2)^2\), which simplifies to \(3x + 4 = 2x + 4 + 4\sqrt{2x+4} + 4\).

Step 3: Simplify

The equation simplifies further to \(x = 4\sqrt{2x + 4}\). We then square the sides again to end up with \(x^2 = 16(2x + 4)\). From here, the equation simplifies to \(x^2=32x+64\), which gives the quadratic equation \(x^2-32x+64=0\). Solving this quadratic equation gives two possibilities for the solution.

Step 4: Solve the quadratic equation

The quadratic equation can be solved by using the quadratic formula \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). Substituting the values gives \(x=16\pm\sqrt{256-256}\), yielding \(x=16\).

Step 5: Check if the solution is valid

Substitute the solution \(x=16\) back into the original equation to verify. The left-hand side becomes \(\sqrt{3*16+4} = \sqrt{52}\). The right-hand side is \(\sqrt{2*16 + 4} + 2 = \sqrt{36} = 6\). As \(\sqrt{52}\) is not equal to \(6\), the solution is invalid.