If 2-3i is a root of z^3-7z^2+25z-39=0, find the other two roots?
2 Answers
2+3i, and 3.
Explanation:
I polynomial equations of order 3 (cubics like this) written in the form
Now because
We know
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