Question

Solve the equation. Check for extraneous solutions. $$x=\sqrt{35+2 x}$$

Short answer

The solution to the equation is \(x = 7\). The value \(x = -5\) is an extraneous solution and hence rejected.

Step-by-step solution

Square both sides

Firstly, to get rid of the square root, square both sides of the equation: \(x^2 = (\sqrt{35 + 2x})^2\). This gives us: \(x^2 = 35 + 2x\).

Rearrange to form a quadratic equation

Rearrange the equation to get x terms together and have it equal to 0. Do this by subtracting \(2x\) and 35 from both sides, which results in the equation: \( x^2 - 2x - 35 = 0\).

Factor the quadratic equation

Look for two numbers that add up to -2 and multiply to -35. The numbers -7 and 5 fit this condition. Therefore, the factored form of the equation is: \( (x-7)(x+5) = 0\).

Solve for x

Set each binomial term equal to 0 and solve for x. \(x-7 = 0\) gives \(x = 7\), and \(x+5 = 0\) gives \(x = -5\).

Check for extraneous solutions

Substitute both solutions back into the original equation \(x = \sqrt{35 + 2x}\) to verify if any of them are extraneous solutions. Substituting \(x=7\): \(7 = \sqrt{35 + 2*7}\) simplifies to \(7 = 7\) which is true. Substituting \(x=-5\): \(-5 = \sqrt{35 + 2*-5}\) simplifies to \(-5 = 5\), which is false. Therefore, \(x=-5\) is an extraneous solution.