Question

What is the quadratic equation in standard form with rational coefficents that has a root of 5+4i?

Answer 1

x^2 -10x + 41 = 0

Explanation:

They gave you a root. A root is a solution, is a zero, is an x-intercept(if real).

This is a working backwards problem. They gave you an answer and want the equation back again.

x = 5 + 4i

working backwards:

x - (5 + 4i) = 0

x - 5 - 4i = 0

This left side is a factor of your equation.

If 5+4i is a root then 5-4i is also a root. So do the same thing:

x = 5 - 4i

x - (5 - 4i) = 0

x - 5 + 4i = 0

x - 5 + 4i and

x - 5 - 4i are the factors. Multiply them together to find the quadratic equation. Use distributive property or difference of squares or an area model to model. See image.

Answer 2

x² - 10x + 41 = 0

Explanation:

complex roots occur as a conjugate pair.

given x = 5 + 4i , then x = 5 - 4i is also a root

the corresponding factors are

(x - (5 + 4i) ) , (x - (5 - 4i) ) , that is

(x - 5 - 4i) and (x -5 + 4i)

the quadratic equation is then the product of the factors, that is

(x - 5 - 4i)(x - 5 + 4i) = 0

multiply each term in the second factor by each term in the first factor.

x² - 5x + 4ix - 5x + 25 - 20i - 4ix + 20i - 16i² = 0 ← collect like terms

x² - 10x + 25 - 16i² = 0 [ note i² = - 1 ]

x² - 10x + 25 + 16 = 0

x² - 10x + 41 = 0 ← in standard form