- #1

DragonMaths

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**One of the solutions to**

x

x

^{4}-2x^{3}+kx^{2}+px+36 = 0 is x = 3

*i*

**Prove that this polynomial has no real solutions (roots) and find the real values of k and p.**

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So far the only progress I seem to have made was possibly finding the value of p.

Since x - 3i is a solution , we can say that x + 3i is also a solution.

I then proceeded to replacing x with 3i on one side, and then setting the solution to itself but with x as -3i , as follows :

x

^{4}-2x

^{3}+kx

^{2}+px+36 = x

^{4}-2x

^{3}+kx

^{2}+px+36

(3i)

^{4}-2(3i)

^{3}+k(3i)

^{2}+p(3i)+36 = (-3i)

^{4}-2(-3i)

^{3}+k(-3i)

^{2}+p(-3i)+36

81i

^{4}-54i

^{3}+9ki+3pi+36 = 81i

^{4}+54i

^{3}+9ki-3pi+36

0 = 108i

^{3}-6pi

0=-108i-6pi

0=-(108-6p)i

0=108-6p

6p=108

p=18

This is as much progress as I have made , and I cannot find anything that can help me

Keep in mind this is asked pre-calculus on on 11th grade Advanced Math course with university grade math problems and so on.Any help would be grately appreciated as I have no idea how to continue at this moment.

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