# Solved eq. x^3-x^2+8x+10=0 and had solutions 3i+1, 1-3i is correct?

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We recall the property of complex roots: If a complex number is the root of an equation, then it's conjugate is also the root of the equation.

Fot the beginning, you'v get as roots 2 complex numbers: z and z'(z' is the conjugate of z).

z = a + bi

z' = a - bi

Now,l we'll verify if the complex numbers are the roots of the equation by substituting them into the original equation.

(1+3i)^3 - (1+3i)^2 + 8(1+3i) + 10 = 0

We'll expand the cube using the formula:

(a+b)^3 = a^3 + b^3 + 3ab(a+b)

a = 1 and b = 3i

(1+3i)^3 = 1^3 + (3i)^3 + 3*1*3i*(1+3i)

(1+3i)^3 = 1 - 27i + 9i(1+3i)

We'll remove the brackets:

(1+3i)^3 = 1 - 27i + 9i - 27

We'll combine real parts and imaginary parts:

(1+3i)^3 = -26 + i(9-27)

(1+3i)^3 = -26 - 18i

We'll expand the square using the formula:

(a+b)^2 = a^2 + 2ab + b^2

(1+3i)^2 = 1^2 + 2*1*3i + (3i)^2

(1+3i)^2 = 1 + 6i - 9

(1+3i)^2 = -8 + 6i

We'll substitute the results into the expression (1+3i)^3 - (1+3i)^2 + 8(1+3i) + 10 = 0.

-26 - 18i - (-8 + 6i) + 8 + 24i + 10 = 0

We'll combine like terms:

(-26 + 8 + 8 + 10) + i(-18 - 6 + 24) = 0

0 + 0*i = 0

It is obvious that 1 + 3i is the root of the equation.

According to the rule, the conjugate of 1 + 3i, namely 1 - 3i, is also the root of the equation. So it is not necessary to verify if 1 - 3i is the root, since we've demonstrated that 1 + 3i is the root of the equation.

Conclusion: 1 + 3i and 1 - 3i are the roots of the equation

x^3-x^2+8x+10=0

Note: In calculus, we've substituted i^2 by it's value, -1.

neela | High School Teacher | (Level 3) Valedictorian

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To check  whether  3i+1and 1-3i  are the  the solution of the equation,

of x^3-x^2+8x+10 = 0.

We know that if a+bi is a solution, then the conjugate a-bi is also a solution and there has to be a real root for a cubic. So let c be the real root.

We presume that 1+3i and 1-3i are the roots. Then the sum of the roots = - (-1) = 1.

Therefore 1+3i+1-3i+c =1

2+c =1.

So c = -1.

So  if 1+3i 1nd 1-3i are solutions, then their product is (1+3i)(1-3i) = 1-9(i^2) = 1+9 = 10. So the other root is  = -1. As product of  all 3 roots (1+3i)(1-3i)c  = 10c = -10/1 as per the relation between the roots and coefficients. So c= -1.

We check now whether -1 is a root.

So f(-1) = (-1)^3-(1)^2 +8(-1)+10 = -1-1-8+10 = 0.

Also from the relation between the roots and coefficients, the sum of the roots = - coefficient of x^2/coefficient of x^3 = -(-1) =1.

That proves that 1+3i and 1-3i are the roots of x^3-x^2+8x+10 = 0