The sum of all the roots of the equation |x² - 8x + 15| - 2x + 7 = 0 is

Q: Wait, why do we reject the negative root here?

In modulus equations, we don't just solve the resulting quadratic; we must respect the 'domain' of the case we assumed. If we assume $x$ is in a region where the expression is positive, any root we find that falls outside that region is mathematically extraneous. It's like checking your work against the rules of the game you set at the start!

Q: Can't we just square both sides to get rid of the modulus?

Squaring both sides is a valid technique, but it often introduces 'ghost' roots that don't satisfy the original equation. It also makes the algebra much more complex (turning a quadratic into a quartic). The case-based approach is safer, more logical, and keeps the algebra manageable.

Q: How do I know which case to use?

The modulus function $|f(x)|$ is defined as $f(x)$ if $f(x) \ge 0$ and $-f(x)$ if $f(x) < 0$. You simply find where $f(x) = 0$ to find the 'switching' points. Everything to the left and right of these points is one case, and everything in between is the other. It's all about partitioning the number line.

Question

The sum of all the roots of the equation $|x^2 - 8x + 15| - 2x + 7 = 0$ is

Options:

(A) $9 - \sqrt{3}$

(B) $9 + \sqrt{3}$

(C) $11 - \sqrt{3}$

(D) $11 + \sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at the parabola $y = x^2 - 8x + 15$. It is a classic, opening upwards, crossing the x-axis at $x = 3$ and $x = 5$.

Now, introduce the modulus operator. It acts like a cosmic mirror, taking the portion of the parabola that dips below the x-axis and flipping it upwards. This creates a sharp, 'bumpy' graph.

Our mission is to find where this graph intersects the line $y = 2x - 7$. This is not just algebra; it is a dance between a curve and a line.

### The Critical Gates

Before we dive into the math, we must identify the 'gates' of our problem. The quadratic $x^2 - 8x + 15$ factors beautifully into $(x - 3)(x - 5)$.

These roots, $x = 3$ and $x = 5$, are the critical points. They divide our world into three regions: the left ($x \le 3$), the middle ($3 < x < 5$), and the right ($x \ge 5$).

In the left and right regions, the parabola is already positive, so the modulus does nothing. In the middle, the parabola is negative, so the modulus flips it.

### Case Study: The Outer Regions

Let us tackle the first scenario: $x \le 3$ or $x \ge 5$. Here, the modulus is redundant. Our equation simplifies to:

$$x^2 - 8x + 15 = 2x - 7$$

Rearranging this, we get the quadratic $x^2 - 10x + 22 = 0$. Using the quadratic formula:

$$x = \frac{10 \pm \sqrt{100 - 88}}{2} = 5 \pm \sqrt{3}$$

Now, the moment of truth: validation. We assumed $x \le 3$ or $x \ge 5$. Since $5 + \sqrt{3} \approx 6.73$, it fits the $x \ge 5$ condition perfectly.

However, $5 - \sqrt{3} \approx 3.27$ falls right into the forbidden middle zone. We must reject it. Our survivor from Case I is $x = 5 + \sqrt{3}$.

### The Middle Ground

Now, consider the middle region: $3 < x < 5$. Here, the expression inside the modulus is negative. To satisfy the modulus, we must negate the entire expression:

$$-(x^2 - 8x + 15) = 2x - 7$$

This simplifies to $-x^2 + 8x - 15 = 2x - 7$, which rearranges to $x^2 - 6x + 8 = 0$. Factoring this is a joy:

$$(x - 4)(x - 2) = 0$$

This gives us $x = 4$ and $x = 2$. Again, we check our boundaries. We are in the region $3 < x < 5$.

The value $x = 4$ is a perfect fit. The value $x = 2$ is outside our region, so we discard it. Our survivor from Case II is $x = 4$.

### The Final Summation

We have navigated the traps and found our two valid roots: $x_1 = 5 + \sqrt{3}$ and $x_2 = 4$.

The problem asks for the sum of all roots. Adding them together, we get:

$$(5 + \sqrt{3}) + 4 = 9 + \sqrt{3}$$

It is elegant, it is precise, and it is the result of careful, logical partitioning. The final answer is **$9 + \sqrt{3}$**.

Animated Solution for Mathematics - Quadratic Equations: The sum of all the roots of the equation $∣ x^{2} - 8 x + 15∣ - 2 x + 7 = 0$ is

Select Answer:

Visualized Solution

Understanding the Equation

Critical Points of the Quadratic

Graphing the Modulus Function

Case 1: $x \leq 3$ or $x \geq 5$

Solving Case 1

Validating Case 1 Roots

Case 2: $3 < x < 5$

Solving Case 2

Validating Case 2 Roots

Summing the Valid Roots

The Sigma Insight: Solution of Quadratic Equations

Analyzing the Setup

The Critical Gates

Case Study

The Middle Ground

The Final Summation

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of all the roots of the equation ∣x2−8x+15∣−2x+7=0 is