Question

Solve the equation. Check for extraneous solutions. $$\sqrt{110-x}=x$$

Short answer

The solution to the equation is \(x = 10\).

Step-by-step solution

Squaring both sides

To eliminate the square root, we should square both sides of the equation. By doing so, we get the equation: \((\sqrt{110-x})^2 = x^2\). Simplifying this produces the equation: \(110-x = x^2\).

Rewrite as a quadratic equation

We can rewrite the equation as a quadratic equation in the standard form \(ax^2 + bx + c = 0\). Here, rearranging the equation gives: \(x^2 + x - 110 = 0\).

Solve the quadratic equation

We can now solve for \(x\) using the quadratic formula: \(x = [-b \pm \sqrt{(b^2 - 4ac)}] / (2a)\). Substituting \(a=1\), \(b=1\) and \(c=-110\) into the formula, we find two potential solutions: \(x = -11\) and \(x = 10\).

Check for extraneous solutions

We now substitute each of these potential solutions back into the original equation to verify if they are valid. When \(x = -11\), we obtain \(\sqrt{110 - (-11)} = -11\), which simplifies to \(\sqrt{121} = -11\). This is not true and thus \(x = -11\) is an extraneous solution. When \(x = 10\), we obtain \(\sqrt{110 - 10} = 10\), which simplifies to \(\sqrt{100} = 10\). This is true and thus \(x = 10\) is a valid solution.