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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!

Respuesta :

Answer:

1. There will be two roots not only one

2.a([tex]x^{2}[/tex] +16) =0

Step-by-step explanation:

Simply assume a quadratic equation [tex]ax^{2} +bx+c[/tex]

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[tex]x^{2}[/tex]+c=a[tex]x^{2}[/tex]+16a

Quadratic is a([tex]x^{2}[/tex] +16) =0 here a is any real number.

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