Astudent 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.

explain your answer. you!

1. There will be two roots not only one

2.a( +16) =0

Step-by-step explanation:

Simply assume a quadratic equation

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+c=a+16a

Quadratic is a( +16) =0 here a is any real number.

2 5/8 is the final answer! : )

b. would be the answer : )

step-by-step explanation:

step 1: add -16x to both sides.

16x−4y+−16x=64+−16x  

−4y=−16x+64

step 2: divide both sides by -4.

−4y /-4 = -16x + 64/-4

answer:   y=4x−16

hope i'm right