which equation has only one real solution?

a. (x^2 - 2x + 3 = 3 - 2x)

b. (x^2 + 5x - 2 = 5x + 2)

c. (3x^2 = 14)

d. (2x^2 + 2x - 1 = 4x - 2)

Question

which equation has only one real solution?

a. (x^2 - 2x + 3 = 3 - 2x)

b. (x^2 + 5x - 2 = 5x + 2)

c. (3x^2 = 14)

d. (2x^2 + 2x - 1 = 4x - 2)

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Factoring Quadratics", "Quadratic Formula Application" ], "new_concepts": [ "Number of Real Solutions", "Discriminant of Quadratic" ], "current_concepts": [ "Factoring Quadratics", "Quadratic Formula Application", "Number of Real Solutions", "Discriminant of Quadratic" ] } ## Simplify each equation to standard form $ax^2 + bx + c = 0$ $$ \begin{aligned} &\text{A. } x^2 - 2x + 3 = 3 - 2x \implies x^2 = 0 \ &\text{B. } x^2 + 5x - 2 = 5x + 2 \implies x^2 - 4 = 0 \ &\text{C. } 3x^2 = 14 \implies 3x^2 - 14 = 0 \ &\text{D. } 2x^2 + 2x - 1 = 4x - 2 \implies 2x^2 - 2x + 1 = 0 \end{aligned} $$ ## Determine the number of real solutions for each simplified equation $$ \begin{aligned} &\text{A. } x^2 = 0 \implies x = 0 \quad (\text{exactly one real solution}) \ &\text{B. } x^2 = 4 \implies x = \pm 2 \quad (\text{two real solutions}) \ &\text{C. } x^2 = \frac{14}{3} \implies x = \pm \sqrt{\frac{14}{3}} \quad (\text{two real solutions}) \ &\text{D. } D = (-2)^2 - 4(2)(1) = 4 - 8 = -4 < 0 \quad (\text{no real solutions}) \end{aligned} $$ ## Identify the correct option $$ \text{Equation A has only one real solution.} $$

(A) $x^2 - 2x + 3 = 3 - 2x$

(B) $x^2 + 5x - 2 = 5x + 2$

(C) $3x^2 = 14$

(D) $2x^2 + 2x - 1 = 4x - 2$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Number of Real Solutions" ] }

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QUESTION IMAGE

Question

Explanation:

Simplify each equation to standard form \(ax^2 + bx + c = 0\)

Determine the number of real solutions for each simplified equation

Identify the correct option

Answer: