https://file.aiquickdraw.com/mathbot/file/1727327987513h6szsv40.jpg

The problem in the image asks the following:

**Problem:**  
Given the equations \( x^3 - 4x^2 + x + 6 = 0 \) and \( x^3 - 3x^2 - 4x + k = 0 \), both of which have a common root, what could be the value of \( k \)?

### Solution:

Let the common root be \( r \). Since \( r \) is a root of both equations, we can substitute \( r \) into each of them.

1. From the first equation:  
   \[
   r^3 - 4r^2 + r + 6 = 0
   \]
2. From the second equation:  
   \[
   r^3 - 3r^2 - 4r + k = 0
   \]

We now have two expressions for \( r \):

- \( r^3 - 4r^2 + r + 6 = 0 \)
- \( r^3 - 3r^2 - 4r + k = 0 \)

### Subtract the two equations:

\[
(r^3 - 3r^2 - 4r + k) - (r^3 - 4r^2 + r + 6) = 0
\]

Simplify:

\[
r^3 - 3r^2 - 4r + k - (r^3 - 4r^2 + r + 6) = 0
\]

\[
r^3 - 3r^2 - 4r + k - r^3 + 4r^2 - r - 6 = 0
\]

Cancel out the \( r^3 \) terms:

\[
-3r^2 + 4r^2 - 4r - r + k - 6 = 0
\]

Simplify further:

\[
r^2 - 5r + k - 6 = 0
\]

Thus, we have:

\[
k = 6 - r^2 + 5r
\]

Now, we need to substitute values of \( r \) and check the possible values of \( k \). We will test integer values of \( r \) to match the options.

- **Test \( r = 1 \):**
  \[
  k = 6 - (1)^2 + 5(1) = 6 - 1 + 5 = 10
  \]
  \( k = 10 \), not a valid option.

- **Test \( r = -1 \):**
  \[
  k = 6 - (-1)^2 + 5(-1) = 6 - 1 - 5 = 0
  \]
  \( k = 0 \), not a valid option.

- **Test \( r = 2 \):**
  \[
  k = 6 - (2)^2 + 5(2) = 6 - 4 + 10 = 12
  \]
  \( k = 12 \), which **is** one of the given options.

Thus, the correct value of \( k \) is **12**.

### Final Answer:
**(C) 12**

---

Let me know if you'd like more details on any step or have any questions! Here are 5 related questions to practice:

1. What is the general approach for solving for common roots between two polynomials?
2. Can the above method be used for any pair of polynomial equations?
3. What happens if no common root exists between two equations?
4. How would you factor \( x^3 - 4x^2 + x + 6 = 0 \) to find its roots?
5. What are the possible real roots of cubic equations?

**Tip:** When solving for common roots in two equations, it's useful to subtract the equations and simplify to find relationships between constants like \( k \).

Given the equations x^3 - 4x^2 + x + 6 = 0 and x^3 - 3x^2 - 4x + k = 0, both of which have a common root, what could be the value of k?

Learn how to solve for the value of k in the system of cubic equations x^3 - 4x^2 + x + 6 = 0 and x^3 - 3x^2 - 4x + k = 0, given they have a common root. Explore the steps to subtract equations and solve for k using integer roots. Suitable for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation