Solve [tex]\((x+3)^2 + (x+3) - 2 = 0\)[/tex].

Let [tex]\( u = x + 3 \)[/tex].

Rewrite the equation in terms of [tex]\( u \)[/tex]:

[tex]\[ (u^2 + u - 2) = 0 \][/tex]

Now solve for [tex]\( u \)[/tex]:

[tex]\[ u^2 + u - 2 = 0 \][/tex]



Answer :

Certainly! Let's solve the equation [tex]\((x + 3)^2 + (x + 3) - 2 = 0\)[/tex] step by step by introducing a substitution.

1. Substitution: Let [tex]\( u = x + 3 \)[/tex].

2. Rewrite the equation: Substitute [tex]\( u \)[/tex] into the given equation:

[tex]\((u)^2 + (u) - 2 = 0\)[/tex]

3. Quadratic equation: Now we have the quadratic equation in terms of [tex]\( u \)[/tex]:

[tex]\( u^2 + u - 2 = 0 \)[/tex].

4. Discriminant method: Solve the quadratic equation using 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].

5. Calculate the discriminant:

[tex]\[ \text{Discriminant} = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot (-2) = 1 + 8 = 9. \][/tex]

6. Find the roots:

[tex]\[ u = \frac{-1 \pm \sqrt{9}}{2 \cdot 1} = \frac{-1 \pm 3}{2}. \][/tex]

7. Calculate the two solutions:

[tex]\[ u_1 = \frac{-1 + 3}{2} = \frac{2}{2} = 1, \][/tex]

[tex]\[ u_2 = \frac{-1 - 3}{2} = \frac{-4}{2} = -2. \][/tex]

8. Back-substitute: Replace [tex]\( u \)[/tex] with [tex]\( x + 3 \)[/tex]:

[tex]\[ u_1 = 1 \implies x + 3 = 1 \implies x = 1 - 3 = -2, \][/tex]

[tex]\[ u_2 = -2 \implies x + 3 = -2 \implies x = -2 - 3 = -5. \][/tex]

So, the solutions to the equation [tex]\((x + 3)^2 + (x + 3) - 2 = 0\)[/tex] are:

[tex]\[ x = -2 \quad \text{and} \quad x = -5. \][/tex]