Question

Solve the equation. Check for extraneous solutions. $$\frac{1}{5} x=\sqrt{x-6}$$

Short answer

The solutions to the equation are \( x = 15 \) and \( x = 10 \).

Step-by-step solution

Square Both Sides

First thing is to square both sides of the equation to eliminate the square root. When this operation is done, obtain a new equation which is then \( \left(\frac{1}{5}x\right)^2 = \left(\sqrt{x-6}\right)^2 \), which simplifies to \( \frac{1}{25}x^2 = {x-6} \).

Bring all terms to one side

Next, bring all terms to one side, resulting in a quadratic equation. This gives \( \frac{1}{25}x^2 - x + 6 = 0 \). To make the computations easier, multiply the whole equation by 25 to get rid of the fraction: \( x^2 - 25x + 150 = 0 \).

Solve the Quadratic Equation

Solutions to a quadratic equation of form \( Dx^2 + Ex + F = 0 \) can be found using the quadratic formula \( x = \frac{-E±\sqrt{E^2-4DF}}{2D} \). Using that formula to our equation \( x = \frac{25±\sqrt{25^2-4\times150}}{2} \) we get \( x_{1,2} = 15, 10 \).

Checking for Extraneous Solutions

Substitute each solution back into the original equation to verify if they are valid or an extraneous solution. Substitution verifies that both 15 and 10 satisfy the original equation, hence both solutions are valid.