what are the solutions of the equation $(2x + 3)^2 + 8(2x + 3) + 11 = 0$? use $u$ substitution and the quadratic formula to solve.

$x = \frac{-4 \pm \sqrt{5}}{2}$

$x = \frac{-7 \pm \sqrt{5}}{2}$

$x = -7$ and $x = -2$

$x = -1$ and $x = 4$

Question

what are the solutions of the equation $(2x + 3)^2 + 8(2x + 3) + 11 = 0$? use $u$ substitution and the quadratic formula to solve.

$x = \frac{-4 \pm \sqrt{5}}{2}$

$x = \frac{-7 \pm \sqrt{5}}{2}$

$x = -7$ and $x = -2$

$x = -1$ and $x = 4$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Quadratic Form Equations" ], "new_concepts": [], "current_concepts": [ "Quadratic Form Equations", "Quadratic Formula" ] } ## Apply u-substitution Using the *Quadratic Form Equations* knowledge point $$ \begin{aligned} &\text{Let } u = 2x + 3.\ &u^2 + 8u + 11 = 0 \end{aligned} $$ ## Solve for u using the quadratic formula Using the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 1$, $b = 8$, and $c = 11$: $$ \begin{aligned} u &= \frac{-8 \pm \sqrt{8^2 - 4(1)(11)}}{2(1)}\ u &= \frac{-8 \pm \sqrt{64 - 44}}{2}\ u &= \frac{-8 \pm \sqrt{20}}{2}\ u &= \frac{-8 \pm 2\sqrt{5}}{2}\ u &= -4 \pm \sqrt{5} \end{aligned} $$ ## Back-substitute to solve for x Substitute $u = 2x + 3$ back into the equation: $$ \begin{aligned} 2x + 3 &= -4 \pm \sqrt{5}\ 2x &= -7 \pm \sqrt{5}\ x &= \frac{-7 \pm \sqrt{5}}{2} \end{aligned} $$

(A) $x = \frac{-4 \pm \sqrt{5}}{2}$

(B) $x = \frac{-7 \pm \sqrt{5}}{2}$

(C) $x = -7 \text{ and } x = -2$

(D) $x = -1 \text{ and } x = 4$

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

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QUESTION IMAGE

Question

Explanation:

Apply u-substitution

Solve for u using the quadratic formula

Back-substitute to solve for x

Answer: