Question

A student claims that -4i is the only imaginary root of a quadratic polynomial equation that has real coefficients.

1. What is the student’s mistake?
2. Write one possible polynomial that has the correct roots from part a in standard form.

Please explain your answer. Thank you!

Answer 1

1. There will be two roots not only one

2.a(
`x^(2)` +16) =0

Explanation:

Simply assume a quadratic equation
`ax^(2) +bx+c`

Here a,b,c are real independent variables.

As given by student -4i is one of it's root it must satisfy the equation.

-16a +4ib+c=0

b must be zero otherwise either a or c must be imaginary number which is not possible.

So b =0

then, c= 16a

a
`x^(2)`+c=a
`x^(2)`+16a

Quadratic is a(
`x^(2)` +16) =0 here a is any real number.