Question #117180
Given that 3 + i is a root of the equation z
3 − 3z
2 − 8z + 30 = 0, find the
remaining roots.
1
Expert's answer
2020-05-20T19:54:13-0400

The equation is given as:

z33z28z+30=0z^3-3z^2-8z+30=0

Let's type the equation in the form of:

z36z2+3z2+10z18z+30=0z^3-6z^2+3z^2+10z-18z+30=0

After some substitutions:

z36z2+10z+3z218z+30=0z^3-6z^2+10z+3z^2-18z+30=0

Realease zz from the first three components, and 3 from the remaining ones:

z(z26z+10)+3(z26z+10)=0z(z^2-6z+10)+3(z^2-6z+10)=0

Then, release z26z+10z^2-6z+10 from both multiplications, yields

(z+3)(z26z+10)=0(z+3)(z^2-6z+10)=0

The root of the first bracket is:

z=3z=-3 ;

As for the second bracket, we know that complex roots occur in conjugate pairs:

as 3+i3+i is a root, so must the other be 3i3-i

Now, sum of the roots =3+i+3i=6=3+i+3-i=6 and,

the product of the roots =(3+i)(3i)=93i+3ii2=9+1=10=(3+i)(3-i)=9-3i+3i-i^2=9+1=10

Hence, the roots of the equation are: 3;3i;3+i-3; 3-i; 3+i




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