Let's solve the equation [tex]\((x+3)^2 + (x+3) - 2 = 0\)[/tex] by introducing a substitution. We'll start by letting [tex]\( u = x + 3 \)[/tex].
1. Substitute [tex]\( u \)[/tex] for [tex]\( x + 3 \)[/tex] in the equation:
[tex]\[
u^2 + u - 2 = 0
\][/tex]
2. Now, we need to solve the quadratic equation [tex]\( u^2 + u - 2 = 0 \)[/tex] for [tex]\( u \)[/tex]. We apply the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -2 \)[/tex].
3. Calculate the discriminant:
[tex]\[
\Delta = b^2 - 4ac = 1^2 - 4(1)(-2) = 1 + 8 = 9
\][/tex]
4. Using the quadratic formula:
[tex]\[
u = \frac{-1 \pm \sqrt{9}}{2 \cdot 1} = \frac{-1 \pm 3}{2}
\][/tex]
5. Find the two solutions for [tex]\( u \)[/tex]:
[tex]\[
u_1 = \frac{-1 + 3}{2} = \frac{2}{2} = 1
\][/tex]
[tex]\[
u_2 = \frac{-1 - 3}{2} = \frac{-4}{2} = -2
\][/tex]
6. Now, substitute back [tex]\( u = x + 3 \)[/tex] into each solution to find the corresponding [tex]\( x \)[/tex] values:
- For [tex]\( u_1 = 1 \)[/tex]:
[tex]\[
x + 3 = 1 \quad \Rightarrow \quad x = 1 - 3 = -2
\][/tex]
- For [tex]\( u_2 = -2 \)[/tex]:
[tex]\[
x + 3 = -2 \quad \Rightarrow \quad x = -2 - 3 = -5
\][/tex]
Therefore, the solutions to the equation [tex]\((x + 3)^2 + (x + 3) - 2 = 0\)[/tex] are [tex]\( x = -2 \)[/tex] and [tex]\( x = -5 \)[/tex].
