Question - Solving a Radical Equation
Solution:
Given: $$ x + 5 = \sqrt{x + 35} $$Step 1: Square both sides to eliminate the square root.$$ (x + 5)^2 = (\sqrt{x + 35})^2 $$Step 2: Expand the left side and simplify the right side.$$ x^2 + 10x + 25 = x + 35 $$Step 3: Set the equation to zero and rearrange terms.$$ x^2 + 10x + 25 - x - 35 = 0 $$$$ x^2 + 9x - 10 = 0 $$Step 4: Factor the quadratic equation.$$ (x + 10)(x - 1) = 0 $$Step 5: Solve for x.$$ x + 10 = 0 $$ or $$ x - 1 = 0 $$$$ x = -10 $$ or $$ x = 1 $$Step 6: Check for extraneous solutions by plugging back into the original equation.For $$ x = -10 $$:$$ -10 + 5 \neq \sqrt{-10 + 35} $$$$ -5 \neq 5 $$ (Not a solution)For $$ x = 1 $$:$$ 1 + 5 = \sqrt{1 + 35} $$$$ 6 = \sqrt{36} $$$$ 6 = 6 $$ (Valid solution)Therefore, $$ x = 1 $$ is the solution.
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