find the real solution of the following radical equation.

\\sqrt{y-1} = y-3\

\y = ?\

Question

find the real solution of the following radical equation.

\\sqrt{y-1} = y-3\

\y = ?\

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Radical Equations", "Solving Radical Equations" ], "new_concepts": [ "Extraneous Solutions" ], "current_concepts": [ "Radical Equations", "Solving Radical Equations", "Extraneous Solutions" ] } ## Square both sides of the equation Using the *Solving Radical Equations* knowledge point $$ \begin{aligned} \sqrt{y-1} &= y-3 \ (\sqrt{y-1})^2 &= (y-3)^2 \ y-1 &= y^2 - 6y + 9 \end{aligned} $$ ## Rearrange into standard quadratic form Using the *Solving Radical Equations* knowledge point $$ \begin{aligned} y^2 - 6y - y + 9 + 1 &= 0 \ y^2 - 7y + 10 &= 0 \end{aligned} $$ ## Solve the quadratic equation Using the *Solving Radical Equations* knowledge point $$ \begin{aligned} (y-2)(y-5) &= 0 \ y = 2 \quad &\text{or} \quad y = 5 \end{aligned} $$ ## Check for extraneous solutions We must verify both potential solutions in the original equation $\sqrt{y-1} = y-3$. For $y = 2$: $$ \begin{aligned} \text{LHS} &= \sqrt{2-1} = \sqrt{1} = 1 \ \text{RHS} &= 2-3 = -1 \ 1 & e -1 \quad (\text{Extraneous}) \end{aligned} $$ For $y = 5$: $$ \begin{aligned} \text{LHS} &= \sqrt{5-1} = \sqrt{4} = 2 \ \text{RHS} &= 5-3 = 2 \ 2 &= 2 \quad (\text{Valid}) \end{aligned} $$ Find the real solution of the following radical equation. $$\sqrt{y-1} = y-3$$ $y =$ 5 { "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

Solve the quadratic equation

Check for extraneous solutions

Answer: