Question

A little bit tricky algebra question:

The polynomial
P(x) = 8x^5 - 60x^4 + 126x^3 + ax^2 + bx - 45
has three roots in arithmetic progression, the other two roots are complex of the form ±bi. If b is an interger , find b.

The thing in front of the bi is a plus minus sign.
Thanks

Answer 1

I was thinking in factorize P(x), but thas doesn´t helped.

Answer 2

So you have two equations and three unknowns:

8x^5 - 60x^4 + 126x^3 + ax^2 + bx - 45 <-- equation #1

x=+- bi <-- equation #2

Can you think of a third equation to be able to solve the system?

Answer 3

you need something with a in it....

Answer 4

But bi and -bi are the imaginary roots of p(x), what can we do with x=-+bi
If we plug x=+-bi in P(x), we´ll get 0.

Also the real roots are in arithmetic progression, that means, they ahave the form
x_1
x_1+r
x_2+2r
Where r is the comon difference.

And that is all the information that we have.
What we can do?

Answer 5

I dunno if this is right.\[ax^2 = a(\pm{bi})^2\]\[ax^2=ab^2i^2\]We know: \[i^2=-1\]\[ax^2=-ab^2 \rightarrow \frac{ ax^2 }{ -a }=\frac{ -ab^2 }{ -a } \rightarrow x^2=b^2\]\[\sqrt{b^2}=\pm \sqrt{x^2} \rightarrow b=\pm \sqrt{x}\]

Answer 6

\[b=\pm{x}\]

Answer 7

@Juarismi the approach you have on the attached image is correct.
let the three real roots be
\[n,n+d,n-d\]
and the imaginary roots are: \[\pm ki\]

Answer 8

expand the polynomial, compare the co-efficients

Answer 9

\[
p(x)=(x-n+d)(x-n)(n-n-d)(x-ki)(x^2+k^2)
\]

Answer 10

\[
p(x)=[x^2+(-2n+d)x+n(n-d)](x-n-d)(x^2+k^2)\\
p(x)=[x^3-2nx^2+dx^2+n(n-d)x-nx^2-dx^2-(n+d)(-2n+d)x-\\n(n-d)(n+d)](x^2+k^2)
\]

Answer 11

but one detail:
3 real and 2 complex or imaginary?

complex number = z = a+bi
imaginary number = bi

they are different

Answer 12

Well, then the two roots are "imaginary", they have the form +-bi, where b is an interger and i is the imaginary unit.

Now ,i think, the next we should do is , to solve the two expressions for p(x) that we have and find any relation between the roots, but this seems so tricky.

It may be any easiest way.
What do you think.

Answer 13

no. the tough part is expanding that huge thing.
finding the numbers for "n", d" and "k" aint that bad

Answer 14

news?

Answer 15

you'd have 3 simultaneous equations using three constant co-efficients

Answer 16

Im trying to expand that last expression, for then to compare with the original factorized p(x), and see whats happens.

It´s easy to make a mistake.

I think that take this as sistem of trhee ecuations, is even more complicated, considering that one of those ecuations is a 5th grade ecuation, and that is almost imposible to solve, becuse whe have three variables on it