find the real solution of the following radical equation.

\b - 10 = \sqrt{20 - 2b}\

\b = ?\

Question

find the real solution of the following radical equation.

\b - 10 = \sqrt{20 - 2b}\

\b = ?\

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 $$ \begin{aligned} (b - 10)^2 &= (\sqrt{20 - 2b})^2 \ b^2 - 20b + 100 &= 20 - 2b \end{aligned} $$ ## Rearrange into standard quadratic form $$ \begin{aligned} b^2 - 18b + 80 = 0 \end{aligned} $$ ## Factor and solve for b, then check for extraneous solutions $$ \begin{aligned} (b - 8)(b - 10) &= 0 \ b = 8 \quad &\text{or} \quad b = 10 \end{aligned} $$ Checking $b = 8$: $$ \begin{aligned} 8 - 10 &\stackrel{?}{=} \sqrt{20 - 2(8)} \ -2 & e \sqrt{4} \quad (\text{extraneous}) \end{aligned} $$ Checking $b = 10$: $$ \begin{aligned} 10 - 10 &\stackrel{?}{=} \sqrt{20 - 2(10)} \ 0 &= \sqrt{0} \quad (\text{valid}) \end{aligned} $$ Find the real solution of the following radical equation. $b - 10 = \sqrt{20 - 2b}$ $b =$ 10 { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Algebra", "Solving Radical Equations" ] }

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QUESTION IMAGE

Question

Explanation:

Square both sides of the equation

Rearrange into standard quadratic form

Factor and solve for b, then check for extraneous solutions

Answer: