Question

Solve the equation. Check for extraneous solutions. $$x=\sqrt{30-x}$$

Short answer

The solution to the given equation is \(x = 5\). The solution \(x = -6\) is extraneous and not valid.

Step-by-step solution

Square both sides of the equation

To eliminate the square root, square both sides of the equation: \(x^2 = (30-x)\)

Rearrange the equation to standard form

Rearrange the equation to follow the standard form of a quadratic equation: \(x^2 + x - 30 = 0\)

Factorize the equation

Factorize the quadratic equation and solve for x: \((x - 5)(x + 6) = 0\). This gives us two solutions: \(x = 5\) and \(x = -6\)

Verification of solutions

Substitute the values of x into the original equation to check for any extraneous solutions. Substituting \(x = 5\) results in: \(5 = \sqrt{30-5}\) which simplifies to \(5 = \sqrt{25}\), hence \(5 = 5\) which means that \(x = 5\) is a valid solution. Substituting \(x = -6\) results in: \(-6 = \sqrt{30-(-6)}\), which simplifies to \(-6 = \sqrt{36}\). Because the square root is always positive, there is no real number whose square root is -6, and therefore \(x = -6\) is an extraneous solution and not valid.