Q.Solve the following quadratic equation : $(2+i)x^2-(5-i)x+2(1-i)=0$

For $(2+i)x^2-(5-i)x+2(1-i)=0$: $a=2+i,\ b=-(5-i),\ c=2(1-i)=2-2i$. Rather than expand the discriminant from scratch, try the sum/product check: sum of roots $=\\dfrac{5-i}{2+i}$ and product $=\\dfrac{2-2i}{2+i}$. Testing the candidate roots $x=1-i$ and $x=\\dfrac45-\\dfrac25i$: sum $=1-i+\\dfrac45-\\dfrac25i=\\dfrac95-\\dfrac75i$; multiplying by $(2+i)$ should recover $5-i$: $(2+i)\\left(\\dfrac95-\\dfrac75i\\right)=\\dfrac{18}{5}-\\dfrac{14}{5}i+\\dfrac95i-\\dfrac75i^2=\\dfrac{18}{5}+\\dfrac75-\\dfrac{14}{5}i+\\dfrac95i=\\dfrac{25}{5}-\\dfrac55i=5-i$ (matches). Product check: $(1-i)\left(\dfrac45-\dfrac25i\right)=\dfrac45-\dfrac25i-\dfrac45i+\dfrac25i^2=\ …

With a complex leading coefficient, verify candidate roots by checking that $a\times(\text{sum of roots})=-b$ and $a\times(\text{product of ro …

• Forgetting the leading coefficient $a=2+i$ is not $1$, so sum/product formulas need the extra factor of $ …