find the real solution of the following radical equation.

\x - 1 = \sqrt{2x + 13}\

\x = ?\

Question

find the real solution of the following radical equation.

\x - 1 = \sqrt{2x + 13}\

\x = ?\

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Radical Equations", "Solving Radical Equations", "Extraneous Solutions" ], "new_concepts": [], "current_concepts": [ "Radical Equations", "Solving Radical Equations", "Extraneous Solutions" ] } ## Square both sides of the equation $$ \begin{aligned} (x - 1)^2 &= (\sqrt{2x + 13})^2 \ x^2 - 2x + 1 &= 2x + 13 \end{aligned} $$ ## Solve the quadratic equation $$ \begin{aligned} x^2 - 4x - 12 &= 0 \ (x - 6)(x + 2) &= 0 \ x = 6 \quad &\text{or} \quad x = -2 \end{aligned} $$ ## Check for extraneous solutions $$ \begin{aligned} \text{For } x = 6: \quad & 6 - 1 = \sqrt{2(6) + 13} \implies 5 = \sqrt{25} \quad (\text{True}) \ \text{For } x = -2: \quad & -2 - 1 = \sqrt{2(-2) + 13} \implies -3 = \sqrt{9} \quad (\text{False}) \end{aligned} $$ Find the real solution of the following radical equation. $x - 1 = \sqrt{2x + 13}$ $x =$ 6 { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Algebra", "Radical Equations" ] }

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QUESTION IMAGE

Question

Explanation:

Square both sides of the equation

Solve the quadratic equation

Check for extraneous solutions

Answer: