Let's solve the equation [tex]\((x+3)^2 + (x+3) - 2 = 0\)[/tex].
To simplify our work, let's use substitution. Let [tex]\( u = x + 3 \)[/tex]. Now, we rewrite the original equation in terms of [tex]\( u \)[/tex]:
[tex]\[ u^2 + u - 2 = 0. \][/tex]
Next, we solve the quadratic equation [tex]\( u^2 + u - 2 = 0 \)[/tex]. To do that, we can factorize the quadratic expression:
The quadratic expression can be written in the form:
[tex]\[ u^2 + u - 2 = (u + a)(u + b) = 0, \][/tex]
where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are numbers to be determined.
To find [tex]\( a \)[/tex] and [tex]\( b \)[/tex], we need them to satisfy two conditions:
1. [tex]\( a + b = 1 \)[/tex] (the coefficient of [tex]\( u \)[/tex])
2. [tex]\( ab = -2 \)[/tex] (the constant term)
From these conditions, we can infer:
- The numbers [tex]\( a \)[/tex] and [tex]\( b \)[/tex] should add up to 1.
- The product of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] should be -2.
By trial, error, or intuition:
- Let [tex]\( a = 2 \)[/tex] and [tex]\( b = -1 \)[/tex].
These satisfy the conditions because:
- [tex]\( 2 + (-1) = 1 \)[/tex]
- [tex]\( 2 \times (-1) = -2 \)[/tex]
Thus, we can factorize [tex]\( u^2 + u - 2 \)[/tex] as:
[tex]\[ (u + 2)(u - 1) = 0. \][/tex]
We solve for [tex]\( u \)[/tex]:
[tex]\[ u + 2 = 0 \quad \text{or} \quad u - 1 = 0. \][/tex]
Therefore,
[tex]\[ u = -2 \quad \text{or} \quad u = 1. \][/tex]
Recall that [tex]\( u = x + 3 \)[/tex]. Substitute back to find [tex]\( x \)[/tex]:
1. For [tex]\( u = -2 \)[/tex]:
[tex]\[ x + 3 = -2 \][/tex]
[tex]\[ x = -2 - 3 \][/tex]
[tex]\[ x = -5 \][/tex]
2. For [tex]\( u = 1 \)[/tex]:
[tex]\[ x + 3 = 1 \][/tex]
[tex]\[ x = 1 - 3 \][/tex]
[tex]\[ x = -2 \][/tex]
So, the solutions to the original equation [tex]\((x+3)^2 + (x+3) - 2 = 0\)[/tex] are:
[tex]\[ x = -5 \quad \text{and} \quad x = -2. \][/tex]
Hence, the solutions to the equation are [tex]\( x = -5 \)[/tex] and [tex]\( x = -2 \)[/tex].
