To solve the equation [tex]\((2x + 3)^2 + 8(2x + 3) + 11 = 0\)[/tex], we will use the [tex]\(u\)[/tex]-substitution method and the quadratic formula. Follow these steps:
1. Substitute: Let [tex]\(u = 2x + 3\)[/tex]. This will simplify our equation.
2. Rewrite the equation: With [tex]\(u\)[/tex] substitution, the equation becomes:
[tex]\[u^2 + 8u + 11 = 0\][/tex]
3. Quadratic Formula: We now solve the quadratic equation [tex]\(u^2 + 8u + 11 = 0\)[/tex] using the quadratic formula. [tex]\[u = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\][/tex] where [tex]\(a = 1\)[/tex], [tex]\(b = 8\)[/tex], and [tex]\(c = 11\)[/tex].
4. Calculate the Discriminant: The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[\Delta = b^2 - 4ac = 8^2 - 4 \times 1 \times 11 = 64 - 44 = 20\][/tex]
5. Find the roots of [tex]\(u\)[/tex]: Using the quadratic formula:
[tex]\[
u = \frac{{-8 \pm \sqrt{20}}}{2 \times 1} = \frac{{-8 \pm \sqrt{20}}}{2} = \frac{{-8 \pm 2\sqrt{5}}}{2}
= -4 \pm \sqrt{5}
\][/tex]
Therefore, the two solutions are:
[tex]\[
u_1 = -4 + \sqrt{5} \quad \text{and} \quad u_2 = -4 - \sqrt{5}
\][/tex]
6. Back-Substitute [tex]\(u\)[/tex] to find [tex]\(x\)[/tex]: Recall that [tex]\(u = 2x + 3\)[/tex]. We now solve for [tex]\(x\)[/tex] from each value of [tex]\(u\)[/tex].
- For [tex]\(u_1 = -4 + \sqrt{5}\)[/tex]:
[tex]\[
-4 + \sqrt{5} = 2x + 3 \implies 2x = -4 + \sqrt{5} - 3 \implies 2x = -7 + \sqrt{5} \implies x = \frac{{-7 + \sqrt{5}}}{2}
\][/tex]
- For [tex]\(u_2 = -4 - \sqrt{5}\)[/tex]:
[tex]\[
-4 - \sqrt{5} = 2x + 3 \implies 2x = -4 - \sqrt{5} - 3 \implies 2x = -7 - \sqrt{5} \implies x = \frac{{-7 - \sqrt{5}}}{2}
\][/tex]
7. Solutions: Therefore, the solutions to the equation [tex]\((2x + 3)^2 + 8(2x + 3) + 11 = 0\)[/tex] are:
[tex]\[
x = \frac{{-7 + \sqrt{5}}}{2} \quad \text{and} \quad x = \frac{{-7 - \sqrt{5}}}{2}
\][/tex]
These are the correct solutions, which align with choice:
[tex]\[ x = \frac{{-7 \pm \sqrt{5}}}{2} \][/tex]
