Answer :
To solve the equation [tex]\((x+3)^2 + (x+3) - 2 = 0\)[/tex], let's proceed with the following steps:
1. Substitute [tex]\( u \)[/tex] for [tex]\( x + 3 \)[/tex]:
Let [tex]\( u = x + 3 \)[/tex].
2. Rewrite the equation: When substituted, the equation becomes:
[tex]\[ u^2 + u - 2 = 0 \][/tex]
3. Solve the quadratic equation [tex]\( u^2 + u - 2 = 0 \)[/tex]:
The general form of a quadratic equation is [tex]\( au^2 + bu + c = 0 \)[/tex].
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -2 \)[/tex].
4. Calculate the discriminant ([tex]\( \Delta \)[/tex]):
The discriminant of a quadratic equation [tex]\( au^2 + bu + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in our coefficients, we get:
[tex]\[ \Delta = 1^2 - 4 \cdot 1 \cdot (-2) = 1 + 8 = 9 \][/tex]
5. Find the roots [tex]\( u_1 \)[/tex] and [tex]\( u_2 \)[/tex] using the quadratic formula:
The quadratic formula for solving [tex]\( au^2 + bu + c = 0 \)[/tex] is:
[tex]\[ u = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and the discriminant, we get:
[tex]\[ u = \frac{-1 \pm \sqrt{9}}{2 \cdot 1} = \frac{-1 \pm 3}{2} \][/tex]
So, we have 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. Substitute back to find [tex]\( x \)[/tex]:
Recall that [tex]\( u = x + 3 \)[/tex]. So, for each value of [tex]\( u \)[/tex]:
For [tex]\( u_1 = 1 \)[/tex]:
[tex]\[ 1 = x + 3 \][/tex]
[tex]\[ x = 1 - 3 \][/tex]
[tex]\[ x = -2 \][/tex]
For [tex]\( u_2 = -2 \)[/tex]:
[tex]\[ -2 = x + 3 \][/tex]
[tex]\[ x = -2 - 3 \][/tex]
[tex]\[ x = -5 \][/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].
1. Substitute [tex]\( u \)[/tex] for [tex]\( x + 3 \)[/tex]:
Let [tex]\( u = x + 3 \)[/tex].
2. Rewrite the equation: When substituted, the equation becomes:
[tex]\[ u^2 + u - 2 = 0 \][/tex]
3. Solve the quadratic equation [tex]\( u^2 + u - 2 = 0 \)[/tex]:
The general form of a quadratic equation is [tex]\( au^2 + bu + c = 0 \)[/tex].
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -2 \)[/tex].
4. Calculate the discriminant ([tex]\( \Delta \)[/tex]):
The discriminant of a quadratic equation [tex]\( au^2 + bu + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in our coefficients, we get:
[tex]\[ \Delta = 1^2 - 4 \cdot 1 \cdot (-2) = 1 + 8 = 9 \][/tex]
5. Find the roots [tex]\( u_1 \)[/tex] and [tex]\( u_2 \)[/tex] using the quadratic formula:
The quadratic formula for solving [tex]\( au^2 + bu + c = 0 \)[/tex] is:
[tex]\[ u = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and the discriminant, we get:
[tex]\[ u = \frac{-1 \pm \sqrt{9}}{2 \cdot 1} = \frac{-1 \pm 3}{2} \][/tex]
So, we have 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. Substitute back to find [tex]\( x \)[/tex]:
Recall that [tex]\( u = x + 3 \)[/tex]. So, for each value of [tex]\( u \)[/tex]:
For [tex]\( u_1 = 1 \)[/tex]:
[tex]\[ 1 = x + 3 \][/tex]
[tex]\[ x = 1 - 3 \][/tex]
[tex]\[ x = -2 \][/tex]
For [tex]\( u_2 = -2 \)[/tex]:
[tex]\[ -2 = x + 3 \][/tex]
[tex]\[ x = -2 - 3 \][/tex]
[tex]\[ x = -5 \][/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].
