To solve the equation \(\frac{(x+16)^2}{9} = 7\), we will first eliminate the fraction by multiplying both sides by 9:
\[ (x + 16)^2 = 63 \]
Next, we will take the square root of both sides. Remember to consider both the positive and negative roots:
\[ x + 16 = \pm \sqrt{63} \]
Now, we can simplify \(\sqrt{63}\):
\[ \sqrt{63} = \sqrt{9 \cdot 7} = 3\sqrt{7} \]
So, we have:
\[ x + 16 = 3\sqrt{7} \quad \text{or} \quad x + 16 = -3\sqrt{7} \]
Now, we will solve for \(x\) in both cases:
- For \(x + 16 = 3\sqrt{7}\):
\[ x = 3\sqrt{7} - 16 \]
- For \(x + 16 = -3\sqrt{7}\):
\[ x = -3\sqrt{7} - 16 \]
Next, we will calculate the numerical values for both solutions. First, we find \(\sqrt{7}\):
\(\sqrt{7} \approx 2.6457513110645906\)
Now, substituting that into the solutions:
- Calculating \(x = 3\sqrt{7} - 16\):
\[ x \approx 3(2.6457513110645906) - 16 \] \[ x \approx 7.9372539331937718 - 16 \] \[ x \approx -8.062746066806229 \]
- Calculating \(x = -3\sqrt{7} - 16\):
\[ x \approx -3(2.6457513110645906) - 16 \] \[ x \approx -7.9372539331937718 - 16 \] \[ x \approx -23.93725393319377 \]
Thus, the two solutions rounded to the nearest hundredth are:
\[ x \approx -8.06 \quad \text{and} \quad x \approx -23.94 \]
Final answers:
\[ x \approx -8.06 \quad \text{and} \quad x \approx -23.94 \]