Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.3

Question 4

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Solve complex quadratic equations by completing the square .

(vi) 3z25z+7=0 3z^2 - 5z + 7 = 0

Solution

Divide both sides by 3:z253z+73=0Isolate the constant term:z253z=73Add (5/32)2=(56)2=2536 to both sides:z253z+2536=73+2536(z56)2=8436+2536=5936Take square root:z56=±5936=±i596z=56±i596=5±i596\begin{aligned} & \boxed{\text{Divide both sides by 3:}} \\ \\ & z^2 - \frac{5}{3}z + \frac{7}{3} = 0 \\ \\ & \boxed{\text{Isolate the constant term:}} \\ \\ & z^2 - \frac{5}{3}z = -\frac{7}{3} \\ \\ & \boxed{\text{Add } \left(\frac{-5/3}{2}\right)^2 = \left(-\frac{5}{6}\right)^2 = \frac{25}{36} \text{ to both sides:}} \\ \\ & z^2 - \frac{5}{3}z + \frac{25}{36} = -\frac{7}{3} + \frac{25}{36} \\ \\ & \left(z - \frac{5}{6}\right)^2 = -\frac{84}{36} + \frac{25}{36} = -\frac{59}{36} \\ \\ & \boxed{\text{Take square root:}} \\ \\ & z - \frac{5}{6} = \pm \sqrt{-\frac{59}{36}} = \pm \frac{i\sqrt{59}}{6} \\ \\ & z = \frac{5}{6} \pm \frac{i\sqrt{59}}{6} = \frac{5 \pm i\sqrt{59}}{6} \end{aligned}