Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.3

Question 4

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Solve complex quadratic equations by completing the square .

(v) 2z2+6z+9=0 2z^2 + 6z + 9 = 0

Solution

Divide both sides by 2:z2+3z+92=0Isolate the constant term:z2+3z=92Add (32)2=94 to both sides:z2+3z+94=92+94(z+32)2=184+94=94Take square root:z+32=±94=±32iz=32±32i=3±3i2\begin{aligned} & \boxed{\text{Divide both sides by 2:}} \\ \\ & z^2 + 3z + \frac{9}{2} = 0 \\ \\ & \boxed{\text{Isolate the constant term:}} \\ \\ & z^2 + 3z = -\frac{9}{2} \\ \\ & \boxed{\text{Add } \left(\frac{3}{2}\right)^2 = \frac{9}{4} \text{ to both sides:}} \\ \\ & z^2 + 3z + \frac{9}{4} = -\frac{9}{2} + \frac{9}{4} \\ \\ & \left(z + \frac{3}{2}\right)^2 = -\frac{18}{4} + \frac{9}{4} = -\frac{9}{4} \\ \\ & \boxed{\text{Take square root:}} \\ \\ & z + \frac{3}{2} = \pm \sqrt{-\frac{9}{4}} = \pm \frac{3}{2}i \\ \\ & z = -\frac{3}{2} \pm \frac{3}{2}i = \frac{-3 \pm 3i}{2} \end{aligned}