Complex root used to find $a$.

Question

Find $a$ if $z=2+i$ is a root of $2z^{2}+3z+2a-14+3i=0$

Workings:

$2(2+i)^{2}+3(2+i)+2a-14+3i=0$

$2(4+4i-1)+6+3i+2a-14+3i=0$

$8-8i-2+6+3i+2a-14+3i=0$

$-2+14i=-2a$

$1-7i=a$

Is this correct?

Answer 1

Your answer is correct.

You could say $2a=14-3i-3z-2z^2=14-3i-3(2+i)-2(2+i)^2$

$=14-3i-3(2+i)-2(3+4i)=2-14i,$ so $a=1-7i$.

By Vieta's formulas, the sum of the roots is $-\dfrac32$, so the other root besides $2+i$ is $-\dfrac72-i$.

As noted in comments, these roots are not complex conjugates;

their product is $(2a-14+3i)/2,$ which is not real.