If 2-3i is a root of z^3-7z^2+25z-39=0, find the other two roots?

2 Answers
Jun 6, 2017

2+3i, and 3.

Explanation:

I polynomial equations of order 3 (cubics like this) written in the form ax^3 + bx^2 + cx + d = 0, with three roots, alpha, beta, and gamma, it is a rule that alpha + beta + gamma = -b-:a, also alpha*beta + beta*gamma + gamma*alpha = c-:a, and that alpha*beta*gamma = -d-:a.
Now because d is real and not imaginary, that means alpha*beta*gamma is real. You should know that to get a real number by multiplying imiginary numbers, you times it by it's conjugate, which for 2-3i is 2+3i. So now we know that alpha and beta are 2-3i and 2+3i.
We know alpha*beta*gamma = -d-: a, so (2-3i)(2+3i)gamma = -(-39)-:1. Expanding (2-3i)(2+3i) we get 13, so 13gamma = 39, thus gamma = 39-:13 = 3, therefore giving us the other two roots: 2+3i and 3.

Jun 6, 2017

2+3i" and " 3

Explanation:

"the complex roots of polynomial equations always "
"occur in "color(blue)"conjugate pairs"

2-3i" is a root "rArr2+3i" is also a root"

"the quadratic factor formed by these roots is "

(z-(2-3i))(z-(2+3i))

=((z-2)+3i)((z-2)-3i)

=(z-2)^2-9i^2

=z^2-4z+4+9

=z^2-4z+13

rArrz^3-7z^2+25z-39

=color(red)(z)(z^2-4z+13)color(magenta)(+4z^2)-7z^2color(magenta)(-13z)+25z-39

=color(red)(z)(z^2-4z+13)color(red)(-3)(z^2-4z+13)color(magenta)(-12z)+12zcolor(magenta)(+39)
color(white)(=)-39

=color(red)(z)(z^2-4z+13)color(red)(-3)(z^2-4z+13)+0

rArr(z-3)" is a root"

rArr(z-3)(z-(2-3i))(z-(2+3i))=0

rArr"roots are " z=3" and " z=2+-3i