⑵　 3 次方程式 x3 + ax
2 + 25x + b = 0が2 + 3i を解にもつとき，実数の定数 a，b の値
は，a = - ウ ，b = - エオ であり，2 + 3i 以外の解は，x = カ ， キ - ク i で
ある。ただし，i は虚数単位とする。

Given the cubic equation \( x^3 + ax^2 + 25x + b = 0 \) where \( 2 + 3i \) is one of the roots, we are tasked with finding the values of the constants \( a \) and \( b \) and the other roots of the equation.

### Step 1: Use the complex conjugate root theorem
Since the coefficients of the polynomial are real, the complex conjugate of \( 2 + 3i \), which is \( 2 - 3i \), must also be a root.

### Step 2: Express the cubic polynomial
If \( 2 + 3i \) and \( 2 - 3i \) are roots, then we can express the cubic polynomial as:
\[
(x - (2 + 3i))(x - (2 - 3i))(x - r) = 0
\]
where \( r \) is the remaining real root.

### Step 3: Expand the product of the conjugate pair
First, compute the product of the two conjugate factors:
\[
(x - 2 - 3i)(x - 2 + 3i) = \left(x - 2\right)^2 - (3i)^2 = (x - 2)^2 + 9
\]
\[
= x^2 - 4x + 4 + 9 = x^2 - 4x + 13
\]

### Step 4: Multiply by the remaining root
Now, multiply by \( (x - r) \):
\[
(x^2 - 4x + 13)(x - r) = x^3 - rx^2 - 4x^2 + 4rx + 13x - 13r
\]
\[
= x^3 + (-r - 4)x^2 + (4r + 13)x - 13r
\]

### Step 5: Match coefficients with the original equation
Now, compare this with the original equation \( x^3 + ax^2 + 25x + b = 0 \). This gives us:
\[
a = -(r + 4), \quad 25 = 4r + 13, \quad b = -13r
\]

### Step 6: Solve for \( r \)
From \( 25 = 4r + 13 \):
\[
4r = 12 \quad \Rightarrow \quad r = 3
\]

### Step 7: Determine \( a \) and \( b \)
Substitute \( r = 3 \) into the equations for \( a \) and \( b \):
\[
a = -(3 + 4) = -7
\]
\[
b = -13 \times 3 = -39
\]

### Step 8: Write down the other root
Given that \( r = 3 \), the other root besides \( 2 + 3i \) and \( 2 - 3i \) is \( x = 3 \).

### Conclusion
- The value of \( a \) is \( -7 \).
- The value of \( b \) is \( -39 \).
- The other roots are \( 2 - 3i \) and \( 3 \).

Thus, the values are:
- \( a = -7 \)
- \( b = -39 \)
- The remaining roots are \( x = 3 \) and \( x = 2 - 3i \).

---

Do you want more details or have any questions? Here are some related questions:

1. How do we know the conjugate of a complex root is also a root?
2. What happens if we miscalculate the roots of a polynomial?
3. How do you determine the nature of the roots based on the discriminant of a cubic equation?
4. What is the significance of the complex conjugate root theorem?
5. How can you verify the roots by substituting them back into the original equation?

**Tip:** Always double-check calculations when dealing with complex numbers to avoid errors in the real and imaginary parts.

Learn how to find the values of constants a and b in the cubic equation x^3 + ax^2 + 25x + b = 0 where 2 + 3i is a root. Discover the step-by-step solution and understand the significance of the complex conjugate root theorem. Suitable for high school students studying cubic equations and complex numbers.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation