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Worked solution

Solve two quadratics with nonreal roots

Solve (a) z2−4z+24=0z^2-4z+24=0 and (b) (3z+1)2+4=0(3z+1)^2+4=0.

CJC · Self-Practice 5

(a) Complete the square

Answer

z=2±25 iz=2\pm2\sqrt5\,i.

Complete the square:

z2−4z+24=0,(z−2)2+20=0,z−2=±20 i=±25 i,z=2±25 i.\begin{aligned}z^2-4z+24&=0,\\(z-2)^2+20&=0,\\z-2&=\pm\sqrt{20}\,i=\pm2\sqrt5\,i,\\z&=2\pm2\sqrt5\,i.\end{aligned}

How did this part go?

(b) Undo the scale and translation

Answer

z=−13±23iz=-\dfrac13\pm\dfrac23i.

Rearrange before taking both square roots:

(3z+1)2=−4,3z+1=±2i,z=−1±2i3=−13±23i.\begin{aligned}(3z+1)^2&=-4,\\3z+1&=\pm2i,\\z&=\frac{-1\pm2i}3=-\frac13\pm\frac23i.\end{aligned}

How did this part go?

Solution notes

Both copies print the two final answer pairs. The explanatory derivations here are authored from the printed equations.