use the quadratic formula to solve (2x^2 + 3x = 4). which of the following are the solutions to the nearest hundredth?

a. 2.35 and 0.85
b. 2.35 and -0.85
c. -2.35 and 0.85
d. -2.35 and -0.85

Question

use the quadratic formula to solve (2x^2 + 3x = 4). which of the following are the solutions to the nearest hundredth?

a. 2.35 and 0.85
b. 2.35 and -0.85
c. -2.35 and 0.85
d. -2.35 and -0.85

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Quadratic Formula Application", "Solving Quadratic Equations" ], "new_concepts": [], "current_concepts": [ "Quadratic Formula Application", "Solving Quadratic Equations" ] } ## Rearrange the equation to standard form $$ \begin{aligned} 2x^2 + 3x &= 4 \ 2x^2 + 3x - 4 &= 0 \end{aligned} $$ ## Apply the quadratic formula $$ \begin{aligned} a &= 2, \quad b = 3, \quad c = -4 \ x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \ x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-4)}}{2(2)} \ x &= \frac{-3 \pm \sqrt{9 + 32}}{4} \ x &= \frac{-3 \pm \sqrt{41}}{4} \end{aligned} $$ ## Calculate the decimal approximations $$ \begin{aligned} \sqrt{41} &\approx 6.4031 \ x_1 &= \frac{-3 + 6.4031}{4} \approx \frac{3.4031}{4} \approx 0.85 \ x_2 &= \frac{-3 - 6.4031}{4} \approx \frac{-9.4031}{4} \approx -2.35 \end{aligned} $$

A. 2.35 and 0.85

B. 2.35 and -0.85

C. -2.35 and 0.85

D. -2.35 and -0.85

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Quadratic Formula Application" ] }

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QUESTION IMAGE

Question

Explanation:

Rearrange the equation to standard form

Apply the quadratic formula

Calculate the decimal approximations

Answer: