Final answer:
To solve \(\sqrt{(x-3)}+5=x\), we isolate \(x\), square both sides, obtain a quadratic equation, factor it, and find the solutions \(x = 4\) and \(x = 7\), of which \(x = 4\) is extraneous and \(x = 7\) is the correct solution.
Step-by-step explanation:
To solve the equation \(\sqrt{(x-3)}+5=x\) for x, we will perform operations step by step to isolate x. First, we move all terms containing x to one side and the constant terms to the other side.
Subtract 5 from both sides: \(\sqrt{(x-3)} = x - 5\)Square both sides to eliminate the square root: \((\sqrt{(x-3)})^2 = (x - 5)^2\)Expand the right side using the FOIL method: \(x-3 = x^2 - 10x + 25\)Rearrange the equation to form a quadratic equation: \(0 = x^2 - 11x + 28\)Factor the quadratic equation: \(0 = (x - 4)(x - 7)\)Solve for x by setting each factor equal to zero: Check each potential solution in the original equation to verify whether each or is an extraneous solution.
Plugging x = 4 into the original equation, we get \(\sqrt{(4-3)}+5 = 4\), which simplifies to \(1 + 5 = 4\), which is not true, so x = 4 is an extraneous solution.
Plugging x = 7 into the original equation, we get \(\sqrt{(7-3)}+5 = 7\), which simplifies to \(2 + 5 = 7\), which is true, so x = 7 is the correct solution.