Math Problem Statement

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。

Solution

Given the cubic equation x3+ax2+25x+b=0x^3 + ax^2 + 25x + b = 0 where 2+3i2 + 3i is one of the roots, we are tasked with finding the values of the constants aa and bb 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+3i2 + 3i, which is 23i2 - 3i, must also be a root.

Step 2: Express the cubic polynomial

If 2+3i2 + 3i and 23i2 - 3i are roots, then we can express the cubic polynomial as: (x(2+3i))(x(23i))(xr)=0(x - (2 + 3i))(x - (2 - 3i))(x - r) = 0 where rr is the remaining real root.

Step 3: Expand the product of the conjugate pair

First, compute the product of the two conjugate factors: (x23i)(x2+3i)=(x2)2(3i)2=(x2)2+9(x - 2 - 3i)(x - 2 + 3i) = \left(x - 2\right)^2 - (3i)^2 = (x - 2)^2 + 9 =x24x+4+9=x24x+13= x^2 - 4x + 4 + 9 = x^2 - 4x + 13

Step 4: Multiply by the remaining root

Now, multiply by (xr)(x - r): (x24x+13)(xr)=x3rx24x2+4rx+13x13r(x^2 - 4x + 13)(x - r) = x^3 - rx^2 - 4x^2 + 4rx + 13x - 13r =x3+(r4)x2+(4r+13)x13r= 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 x3+ax2+25x+b=0x^3 + ax^2 + 25x + b = 0. This gives us: a=(r+4),25=4r+13,b=13ra = -(r + 4), \quad 25 = 4r + 13, \quad b = -13r

Step 6: Solve for rr

From 25=4r+1325 = 4r + 13: 4r=12r=34r = 12 \quad \Rightarrow \quad r = 3

Step 7: Determine aa and bb

Substitute r=3r = 3 into the equations for aa and bb: a=(3+4)=7a = -(3 + 4) = -7 b=13×3=39b = -13 \times 3 = -39

Step 8: Write down the other root

Given that r=3r = 3, the other root besides 2+3i2 + 3i and 23i2 - 3i is x=3x = 3.

Conclusion

  • The value of aa is 7-7.
  • The value of bb is 39-39.
  • The other roots are 23i2 - 3i and 33.

Thus, the values are:

  • a=7a = -7
  • b=39b = -39
  • The remaining roots are x=3x = 3 and x=23ix = 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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Cubic Equations
Complex Numbers
Roots of Polynomials

Formulas

Cubic polynomial structure
Complex conjugate root theorem

Theorems

Complex conjugate root theorem

Suitable Grade Level

High School