Question

follow the step-by-step process to solve the equation by using the quadratic formula. complete parts a through c.

(3x^2 + 11x = 20)

a. after writing the given equation in the form (ax^2 + bx + c = 0), where (a \
eq 0), identify the values of (a), (b), and (c).
(a = \box), (b = \box), (c = \box)

b. correctly substitute the values of (a), (b), and (c) into the quadratic formula, (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), and do not simplify. choose the correct answer.

a. (x = \frac{11 \pm \sqrt{(11)^2 + 4(3)(-20)}}{2(3)})
b. (x = \frac{-11 \pm \sqrt{(11)^2 - 4(3)(-20)}}{2(3)})
c. (x = \frac{-11 \pm \sqrt{(11)^2 + 4(3)(-20)}}{2(3)})
d. (x = \frac{11 \pm \sqrt{(11)^2 - 4(3)(-20)}}{2(3)})

c. completely simplify the right side of the quadratic formula from the previous step to solve for (x).
(x = \box)
(simplify your answer. use a comma to separate answers as needed. use integers or fractions for any numbers in the expression. type an exact answer, using radicals and (i) as needed.)

Explanation:

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<reasoning>

Write in standard form and identify coefficients

\[

$$\begin{aligned} &3x^2 + 11x = 20 \\ &3x^2 + 11x - 20 = 0 \\ &a = 3,\quad b = 11,\quad c = -20 \end{aligned}$$

\]

Substitute into the quadratic formula

\[

$$\begin{aligned} &x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &x = \frac{-11 \pm \sqrt{(11)^2 - 4(3)(-20)}}{2(3)} \end{aligned}$$

\]

Simplify to solve for x

\[

$$\begin{aligned} &x = \frac{-11 \pm \sqrt{121 + 240}}{6} \\ &x = \frac{-11 \pm \sqrt{361}}{6} \\ &x = \frac{-11 \pm 19}{6} \\ &x_1 = \frac{-11 + 19}{6} = \frac{8}{6} = \frac{4}{3} \\ &x_2 = \frac{-11 - 19}{6} = \frac{-30}{6} = -5 \end{aligned}$$

\]
</reasoning>

<answer>

Question a

After writing the given equation in the form \(ax^2 + bx + c = 0\), where \(a
eq 0\), identify the values of \(a\), \(b\), and \(c\).
\(a =\) <blank>3</blank>
\(b =\) <blank>11</blank>
\(c =\) <blank>-20</blank>

Question b

<mcq-option>(A) \(x = \frac{11 \pm \sqrt{(11)^2 + 4(3)(-20)}}{2(3)}\)</mcq-option>
<mcq-option>(B) \(x = \frac{-11 \pm \sqrt{(11)^2 - 4(3)(-20)}}{2(3)}\)</mcq-option>
<mcq-correct>(C) \(x = \frac{-11 \pm \sqrt{(11)^2 - 4(3)(-20)}}{2(3)}\)</mcq-correct>
<mcq-option>(D) \(x = \frac{11 \pm \sqrt{(11)^2 - 4(3)(-20)}}{2(3)}\)</mcq-option>

Question c

Completely simplify the right side of the quadratic formula from the previous step to solve for \(x\).
\(x =\) <blank>\(\frac{4}{3}, -5\)</blank>
</answer>

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"Mathematics",
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"Quadratic Equations"
]
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</post_analysis>

Answer:

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<reasoning>

Write in standard form and identify coefficients

\[

$$\begin{aligned} &3x^2 + 11x = 20 \\ &3x^2 + 11x - 20 = 0 \\ &a = 3,\quad b = 11,\quad c = -20 \end{aligned}$$

\]

Substitute into the quadratic formula

\[

$$\begin{aligned} &x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &x = \frac{-11 \pm \sqrt{(11)^2 - 4(3)(-20)}}{2(3)} \end{aligned}$$

\]

Simplify to solve for x

\[

$$\begin{aligned} &x = \frac{-11 \pm \sqrt{121 + 240}}{6} \\ &x = \frac{-11 \pm \sqrt{361}}{6} \\ &x = \frac{-11 \pm 19}{6} \\ &x_1 = \frac{-11 + 19}{6} = \frac{8}{6} = \frac{4}{3} \\ &x_2 = \frac{-11 - 19}{6} = \frac{-30}{6} = -5 \end{aligned}$$

\]
</reasoning>

<answer>

Question a

After writing the given equation in the form \(ax^2 + bx + c = 0\), where \(a
eq 0\), identify the values of \(a\), \(b\), and \(c\).
\(a =\) <blank>3</blank>
\(b =\) <blank>11</blank>
\(c =\) <blank>-20</blank>

Question b

<mcq-option>(A) \(x = \frac{11 \pm \sqrt{(11)^2 + 4(3)(-20)}}{2(3)}\)</mcq-option>
<mcq-option>(B) \(x = \frac{-11 \pm \sqrt{(11)^2 - 4(3)(-20)}}{2(3)}\)</mcq-option>
<mcq-correct>(C) \(x = \frac{-11 \pm \sqrt{(11)^2 - 4(3)(-20)}}{2(3)}\)</mcq-correct>
<mcq-option>(D) \(x = \frac{11 \pm \sqrt{(11)^2 - 4(3)(-20)}}{2(3)}\)</mcq-option>

Question c

Completely simplify the right side of the quadratic formula from the previous step to solve for \(x\).
\(x =\) <blank>\(\frac{4}{3}, -5\)</blank>
</answer>

<post_analysis>
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"question_type": "Multi-part",
"knowledge_point": [
"Mathematics",
"Algebra",
"Quadratic Equations"
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