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 where is one of the roots, we are tasked with finding the values of the constants and 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 , which is , must also be a root.
Step 2: Express the cubic polynomial
If and are roots, then we can express the cubic polynomial as: where is the remaining real root.
Step 3: Expand the product of the conjugate pair
First, compute the product of the two conjugate factors:
Step 4: Multiply by the remaining root
Now, multiply by :
Step 5: Match coefficients with the original equation
Now, compare this with the original equation . This gives us:
Step 6: Solve for
From :
Step 7: Determine and
Substitute into the equations for and :
Step 8: Write down the other root
Given that , the other root besides and is .
Conclusion
- The value of is .
- The value of is .
- The other roots are and .
Thus, the values are:
- The remaining roots are and .
Do you want more details or have any questions? Here are some related questions:
- How do we know the conjugate of a complex root is also a root?
- What happens if we miscalculate the roots of a polynomial?
- How do you determine the nature of the roots based on the discriminant of a cubic equation?
- What is the significance of the complex conjugate root theorem?
- 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.
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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
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