use the zero product property to find the solutions to the equation (2x^2 + x - 1 = 2)

(x = -\frac{1}{2}) or (x = 2)

(x = -2) or (x = \frac{1}{2})

(x = -\frac{3}{2}) or (x = 1)

(x = 1) or (x = \frac{3}{2})

Question

use the zero product property to find the solutions to the equation (2x^2 + x - 1 = 2)

(x = -\frac{1}{2}) or (x = 2)

(x = -2) or (x = \frac{1}{2})

(x = -\frac{3}{2}) or (x = 1)

(x = 1) or (x = \frac{3}{2})

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Zero Product Property Interpretation", "Solving Quadratic Equations" ], "new_concepts": [], "current_concepts": [ "Zero Product Property Interpretation", "Solving Quadratic Equations", "Factoring Quadratics" ] } ## Set the quadratic equation to zero $$ \begin{aligned} 2x^2 + x - 1 &= 2 \ 2x^2 + x - 3 &= 0 \end{aligned} $$ ## Factor the quadratic expression $$ \begin{aligned} 2x^2 + 3x - 2x - 3 &= 0 \ x(2x + 3) - 1(2x + 3) &= 0 \ (2x + 3)(x - 1) &= 0 \end{aligned} $$ ## Apply the zero product property to solve for x $$ \begin{aligned} 2x + 3 = 0 &\implies x = -\frac{3}{2} \ x - 1 = 0 &\implies x = 1 \end{aligned} $$

(A) $x = -\frac{1}{2}$ or $x = 2$

(B) $x = -2$ or $x = \frac{1}{2}$

(C) $x = -\frac{3}{2}$ or $x = 1$

(D) $x = 1$ or $x = \frac{3}{2}$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Solving Quadratic Equations" ] }

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QUESTION IMAGE

Question

Explanation:

Set the quadratic equation to zero

Factor the quadratic expression

Apply the zero product property to solve for x

Answer: