SOLVING EQUATIONS WITH COMPLEX ROOTS

Example 1 :

If (2 + 3i) is one of the roots of the following quadratic equation, find the other root.

x² - 4x + 13 = 0

Solution :

Since (2 + 3i) is a complex root, (2 - 3i) must be the other root.

So, the other root of the given quadratic equation is

2 - 3i

Example 2 :

Solve the following quadratic equation :

x² + 4x + 5 = 0

Solution :

The given quadratic equation can not be solved by factoring. So, we can use quadratic formula.

Comparing ax² + bx + c = 0 and x² + 4x + 5 = 0,

a = 1, b = 4, c = 5

Quadratic formula :

Substitute a = 1, b = 4 and c = 5.

Therefore, the roots of the given quadratic equation are

-2 + i and -2 - i

Example 3 :

Solve the following equation, if one of its root is (1 - i).

x³ - 5x² + 8x - 6 = 0

Solution :

Since (1 - i) is a complex root, (1 + i) must be the other root.

x = 1 - i   or   x - (1 - i) = 0

x = 1 + i   or   x - (1 + i) = 0

Quadratic polynomial with the roots (1 - i) and (1 + i) :

= x² - (sum of the roots)x + product of the roots

= x² - [(1 + i) + (1 - i)]x + (1 + i)(1 - i)

= x² - [1 + i + 1 - i]x + 1² - i²

= x² - 2x + 1 - (-1)

= x² - 2x + 1 + 1

= x² - 2x + 2

(x² - 2x + 2) is a factor of (x³ - 5x² + 8x - 6).

Let (x + a) be the other factor.

Then,

(x + a)(x² - 2x + 2) = (x³ - 5x² + 8x - 6) 

Compare the constant terms.

2a = -6

a = -3

So, (x - 3) is a factor of the given polynomial equation.

x - 3 = 0

x = 3

Therefore, the roots of the given equation are

1 - i, 1 + i and 3

Example 4 :

Solve the following equation, if one of its roots is (2 + i√3)

x⁴ - 4x² + 8x + 35 = 0

Solution :

Since (2 + i√3) is a complex root, (2 - i√3) must be the other root.

x = 2 + i√3   or   x - (2 + i√3) = 0

x = 2 - i√3   or   x - (2 - i√3) = 0

Quadratic polynomial with the roots (2 + i√3) and (2 - i√3) :

= x² - (sum of the roots)x + product of the roots

= x² - [(2 + i√3) + (1 - i2√3)]x + (2 + i√3)(2 - i√3)

= x² - [2 + i√3 + 2 - i√3]x + 2² - (√3i)²

= x² - 4x + 4 - (√3)²i²

= x² - 4x + 4 - 3(-1)

= x² - 4x + 4 + 3

= x² - 4x + 7

(x² - 4x + 7) is a factor of (x⁴ - 4x² + 8x + 35).

Let (x² + ax + b) be the other factor.

Then,

(x² + ax + b)(x² - 4x + 7) = x⁴ - 4x² + 8x + 35

Comparing the constant terms,

7b = 35

b = 5

Comparing the coefficient of x terms,

7a - 4b = 8

Substitute b = 5.

7a - 4(5) = 8

7a - 20 = 8

7a = 28

a = 4

x² + ax + b ----> x² + 4x + 5

Solving the equation x² + 4x + 5 = 0, we get

x = -2 + i or -2 - i

(refer example 1)

Therefore, the roots of the given equation are

2 + i√3, 2 - i√3, -2 + i and -2 - i

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