Question

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

Answer 1

Answer:

1. There will be two roots not only one

2.a($x^{2}$ +16) =0

Step-by-step 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.