Key Concepts and Formulas

• Absolute Value Definition: $|x| = x$ if $x \geq 0$, and $|x| = -x$ if $x < 0$.
• Solving Absolute Value Equations: To solve $|f(x)| = g(x)$, consider two cases: $f(x) = g(x)$ and $f(x) = -g(x)$. Remember to check that $g(x) \geq 0$ for valid solutions.
• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

Step 1: Isolate the Absolute Value We are given the equation $|x^2 - 8x + 15| - 2x + 7 = 0$. Our first step is to isolate the absolute value term on one side of the equation. This allows us to apply the definition of absolute value. $|x^2 - 8x + 15| = 2x - 7$

Step 2: Establish the Non-Negativity Condition Since the absolute value of any expression is non-negative, the right-hand side of the equation must also be non-negative. This gives us a condition on $x$ that any valid root must satisfy. $2x - 7 \geq 0 \Rightarrow 2x \geq 7 \Rightarrow x \geq \frac{7}{2} = 3.5$ This condition, $x \geq 3.5$, will be crucial for eliminating extraneous solutions later.

Step 3: Analyze the Expression Inside the Absolute Value Let $f(x) = x^2 - 8x + 15$. We need to determine the intervals where $f(x) \geq 0$ and $f(x) < 0$. This determines when we use $f(x)$ or $-f(x)$ when removing the absolute value. We can factor the quadratic as $f(x) = (x - 3)(x - 5)$. The roots of $f(x) = 0$ are $x = 3$ and $x = 5$. The parabola opens upwards. Therefore, $f(x) \geq 0$ when $x \leq 3$ or $x \geq 5$, and $f(x) < 0$ when $3 < x < 5$.

Step 4: Case 1: $x^2 - 8x + 15 \geq 0$ (i.e., $x \leq 3$ or $x \geq 5$) In this case, $|x^2 - 8x + 15| = x^2 - 8x + 15$. Substituting this into the equation from Step 1 gives: $x^2 - 8x + 15 = 2x - 7$ $x^2 - 10x + 22 = 0$ We solve this quadratic equation. Using the quadratic formula: $x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(22)}}{2(1)} = \frac{10 \pm \sqrt{100 - 88}}{2} = \frac{10 \pm \sqrt{12}}{2} = \frac{10 \pm 2\sqrt{3}}{2} = 5 \pm \sqrt{3}$ So, we have two potential roots: $x_1 = 5 + \sqrt{3}$ and $x_2 = 5 - \sqrt{3}$.

Step 5: Verify Solutions from Case 1 We must check if these roots satisfy both the condition for Case 1 ($x \leq 3$ or $x \geq 5$) and the non-negativity condition from Step 2 ($x \geq 3.5$).

• For $x_1 = 5 + \sqrt{3} \approx 5 + 1.732 = 6.732$: This satisfies $x \geq 5$ and $x \geq 3.5$. So, $x_1 = 5 + \sqrt{3}$ is a valid root.
• For $x_2 = 5 - \sqrt{3} \approx 5 - 1.732 = 3.268$: This does not satisfy $x \leq 3$ or $x \geq 5$. It also does not satisfy $x \geq 3.5$. So, $x_2 = 5 - \sqrt{3}$ is an extraneous solution.

Step 6: Case 2: $x^2 - 8x + 15 < 0$ (i.e., $3 < x < 5$) In this case, $|x^2 - 8x + 15| = -(x^2 - 8x + 15)$. Substituting this into the equation from Step 1 gives: $-(x^2 - 8x + 15) = 2x - 7$ $-x^2 + 8x - 15 = 2x - 7$ $0 = x^2 - 6x + 8$ $x^2 - 6x + 8 = 0$ Factoring gives $(x - 2)(x - 4) = 0$, so $x = 2$ or $x = 4$.

Step 7: Verify Solutions from Case 2 We must check if these roots satisfy both the condition for Case 2 ($3 < x < 5$) and the non-negativity condition from Step 2 ($x \geq 3.5$).

• For $x_3 = 2$: This does not satisfy $3 < x < 5$. It also does not satisfy $x \geq 3.5$. So, $x_3 = 2$ is an extraneous solution.
• For $x_4 = 4$: This satisfies $3 < x < 5$ and $x \geq 3.5$. So, $x_4 = 4$ is a valid root.

Step 8: Calculate the Sum of All Roots The valid roots are $5 + \sqrt{3}$ and $4$. The sum of the roots is: $4 + (5 + \sqrt{3}) = 9 + \sqrt{3}$

Common Mistakes & Tips

• Forgetting to check the non-negativity condition: The expression on the right-hand side of the isolated absolute value equation must be non-negative. Failure to enforce this condition will lead to extraneous solutions.
• Not verifying roots against case conditions: It is crucial to check each potential solution against the specific interval condition from which it arose. Failing to do so is a very common source of error in absolute value problems.
• Sign errors: Be very careful with signs when substituting $-(x^2 - 8x + 15)$ into the equation.

Summary We solved the given equation by isolating the absolute value, establishing the non-negativity condition $x \geq 3.5$, and then considering two cases based on the sign of the expression inside the absolute value. We found the roots for each case and verified them against the conditions for each case, as well as the non-negativity condition. The sum of the valid roots is $9 + \sqrt{3}$.

The final answer is $\boxed{9+\sqrt{3}}$, which corresponds to option (B).