Find the real solutions of the equation (x + 4)² + 17(x + 4) + 72 = 0.

Step-by-step solution

1. 1

   Notice the repeated block, not the expanded degree. The expression $x+4$ appears squared and then to the first power, and nothing else contains $x$. That is the signature of an equation quadratic in form: expanding to $x^2 + 25x + 156 = 0$ would work, but it throws away the structure that makes the factoring easy.

2. 2

   Substitute $u = x + 4$. Replacing the repeated block turns the equation into a plain quadratic:

   $u^2 + 17u + 72 = 0$

3. 3

   Factor the quadratic in $u$. Look for two numbers whose product is $72$ and whose sum is $17$. Since $8 \cdot 9 = 72$ and $8 + 9 = 17$:

   $(u+8)(u+9) = 0 \quad\Longrightarrow\quad u = -8 \ \text{ or } \ u = -9$

   The discriminant is $17^2 - 4(72) = 1 > 0$, confirming two distinct real roots.

4. 4

   Undo the substitution — this is the step most often skipped. The question asks for $x$, not $u$, so put $u = x+4$ back:

   $x + 4 = -8 \ \Rightarrow \ x = -12, \qquad x + 4 = -9 \ \Rightarrow \ x = -13$

5. 5

   Check both values in the original equation. For $x=-12$: $(-8)^2 + 17(-8) + 72 = 64 - 136 + 72 = 0$. For $x=-13$: $(-9)^2 + 17(-9) + 72 = 81 - 153 + 72 = 0$. Because the substitution is a simple shift (never squaring or taking roots of both sides), no extraneous solutions can appear.