SOLUTION: Question: Equation: x^3 + ax^2 +bx +a = 0 |a,b are real If x = 2+i is a root of the equation, find a and b

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Question 1163497: Question: Equation: x^3 + ax^2 +bx +a = 0 |a,b are real
If x = 2+i is a root of the equation, find a and b

Found 2 solutions by solver91311, ikleyn:
Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

Complex roots always appear in conjugate pairs. So if is a root of the equation, so is

Therefore both and are factors of the cubic polynomial. I leave it as an exercise for the student to verify that the product of these two factors is

Using Polynomial Long Division

In order for to be a factor of the original cubic polynomial, the following facts must hold:

Solve the 2X2 system for and

John

My calculator said it, I believe it, that settles it

Answer by ikleyn(52788) (Show Source):
You can put this solution on YOUR website!
.

Since the coefficients of the polynomial are real numbers and one root is (2+i), the other root is complex conjugate (2-i). Let the third root is x (it is clear that the third root is a real number). Now use the Vieta's theorem. It says that the product of the roots is the constant term with the opposite sign: (2+i)*(2-i)*x = -a, or 5x = -a. (1) The Vieta;s theorem also says that the sum of the three roots is equal to a coefficient at x^2 with the opposite sign: 2+i + 2-i + x = -a, or 4 + x = -a. (2) So, you have this system of two equations (1) and (2). Based on (1), replace "-a" in (2) by 5x. You will get then 4 + x = 5x Hence, x= 1 and a= -5x = -5. To find the coefficient "b", apply the Vieta's theorem again. It says that the coefficient b is equal to the sum of the three in-pairs products of the roots b = (2+i)*(2-i) + 1*(2+i) + 1*(2-i) = 5 + 2+i + 2-i = 5 + 4 = 9. ANSWER. a= -5, b= 9.

Solved.