To solve the equation [tex]\((x-3)^2 + 2(x-3) - 8 = 0\)[/tex], we can use substitution to make the process simpler.
Let's define a new variable [tex]\( u \)[/tex] such that:
[tex]\[ u = x - 3 \][/tex]
Now, substitute [tex]\( u \)[/tex] back into the equation:
[tex]\[ (u)^2 + 2(u) - 8 = 0 \][/tex]
We now have a quadratic equation in terms of [tex]\( u \)[/tex]:
[tex]\[ u^2 + 2u - 8 = 0 \][/tex]
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\( u^2 + 2u - 8 = 0 \)[/tex], the coefficients are:
[tex]\[ a = 1, \quad b = 2, \quad c = -8 \][/tex]
First, calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-8) = 4 + 32 = 36 \][/tex]
Since the discriminant is positive, we have two distinct real solutions. Now, solve for [tex]\( u \)[/tex]:
[tex]\[ u = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-2 \pm \sqrt{36}}{2 \cdot 1} = \frac{-2 \pm 6}{2} \][/tex]
This gives us two potential values for [tex]\( u \)[/tex]:
[tex]\[ u_1 = \frac{-2 + 6}{2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ u_2 = \frac{-2 - 6}{2} = \frac{-8}{2} = -4 \][/tex]
Recalling our substitution [tex]\( u = x - 3 \)[/tex], we substitute back to find [tex]\( x \)[/tex]:
[tex]\[ 2 = x - 3 \implies x = 2 + 3 = 5 \][/tex]
[tex]\[ -4 = x - 3 \implies x = -4 + 3 = -1 \][/tex]
Therefore, the solutions to the equation [tex]\((x-3)^2 + 2(x-3) - 8 = 0\)[/tex] are:
[tex]\[ x = 5 \quad \text{and} \quad x = -1 \][/tex]
Given the options, the correct answer is:
[tex]\[ \boxed{x = -1 \text{ and } x = 5} \][/tex]
