Question

Solve the equation. Check for extraneous solutions. $$x=\sqrt{4 x+45}$$

Short answer

The solution set for the equation \(x=\sqrt{4 x+45}\) is {9} after checking for extraneous solutions.

Step-by-step solution

Isolate the square root

The equation is already in the form where the square root is isolatable, meaning that \(x = \sqrt{4x + 45}\). The square root is already isolated on the right side.

Eliminate the square root

Remove the square root by squaring both sides of the equation: \(x^2 = (4x + 45)\). We get \(x^2 = 4x + 45\).

Solve the quadratic equation

Rearrange the equation to equal zero: \(x^2 - 4x - 45 = 0\). Next, factor the quadratic, which will give: \((x - 9)(x + 5) = 0\). Setting each factor equal to zero yields two potential solutions: \(x = 9\) and \(x = -5\).

Check for extraneous solutions

We must ensure that these are actual solutions. We place them back into the original equation: \(\sqrt{4x + 45}\). When we place \(x = 9\) into the equation we get 9 = \(\sqrt{4(9) + 45}\). Simplifying gives 9 = \(\sqrt{81}\), which means 9 = 9. This is true, so \(x = 9\) is a valid solution. When we place \(x = -5\) into the equation, however, the result is \(-5 = \sqrt{4(-5) + 45}\), which simplifies to \(-5 = \sqrt{-5}\). Because the square root of a negative number is not a real number, \(x = -5\) is an extraneous solution and should be dismissed.