Solve the equation: $z^2 - (7 +6i)z + 4 +22i = 0 $ in $\mathbb{C}$

Question

I am doing some repetition for fun and got stuck on this question:

Solve the equation: $z^2 - (7 +6i)z + 4 +22i = 0 $ in $\mathbb{C}$

This is actually on the chapter of polynomials so I guess that I should start to factor and group the terms to "see" the answer.

I have not had much luck with that approach yet though...

Please help me out here

Answer 1

$$ z=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\Delta=(b^2-4ac)=(7+6i)^2-4(4+22i)=\\=49+36i^2+84i-16-88i=\\49-36+84i-16-88i=\\-3-4i=\\=-4+1-4i\\=(2i)^2+1-2(2i)=\\(2i-1)^2\\$$so $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\z=\frac{-(7+6i)\pm \sqrt{(2i-1)^2}}{2*1}\\z=\frac{-(7+6i)\pm (2i-1)}{2} $$

Answer 2

You will probably need to use sum and product of roots. Letting the roots be $\alpha, \beta$, we have $\alpha + \beta= 7+6i$, $\alpha \beta=4+22i$. Then solve simultaneously to find $\alpha$ and $\beta$.

You might like to look at Vieta's formulae if you're unfamiliar with this.