# Equation.Calculate x for x^3-x^2+8x+10=0 and x1=3i+1?

Asked on by eligia

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to know that the complex solutions come in pairs, hence, since `x = 1 + 3i` is one solution, then `x = 1 - 3i` is the second solution.

You may use the following Vieta's relation to find the next root, such that:

`x_1 + x_2 + x_3 = -(-1)/1 => 1 + 3i + 1 - 3i + x_3 = 1`

Reducing duplicate members yields:

`2 + x_3 = 1 => x_3 = 1 - 2 => x_3 = -1`

Hence, evaluating the solutions to the given cubic equation, using Vieta's relations, yields `x_(1,2) = 1+-3i, x_3 = -1.`

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Since the equation has a complex root, we'll 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.

So, we have as root the number z = a + bi => z' = a - bi is also the root of the equation.

Now, we'll verify if the complex numbers is the root of the equation by substituting it into the original equation.

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.

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