# Question 2, Exercise 1.3

Solutions of Question 2 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Solve the equation by completing square: $z^{2}-6 z+2=0$.

Solution.

\begin{align} & z^2 - 6z + 2 = 0 \\ \implies & z^2 - 2(3)(z)+9-9+2=0 \\ \implies & (z - 3)^2+7= 0 \\ \implies & (z - 3)^2 = 7. \end{align} Take the square root of both sides: \begin{align} &z - 3 = \pm \sqrt{7} \\ \implies &z = 3 \pm \sqrt{7}\end{align} Hence Solutioin set=$\{3 \pm \sqrt{7}\}$.

Solve the equation by completing square: $-\dfrac{1}{2} z^{2}-5 z+2=0$.

Solution.

\begin{align} -\dfrac{1}{2} z^{2} - 5z + 2& = 0 \end{align} Multiply through by $-2$ to eliminate the fraction: \begin{align} z^2 + 10z - 4 &= 0 \\ z^2 + 10z +25-25-4&=0\\ (z + 5)^2 - 29&=0 \end{align} Take the square root of both sides: \begin{align} z + 5 &= \pm \sqrt{29}\\ \implies z &= -5 \pm \sqrt{29} \end{align} Hence solution set$=\{-5 \pm \sqrt{29}\}$

Solve the equation by completing square: $4 z^{2}+5 z=14$.

Solution.

\begin{align} 4z^{2} + 5z &= 14\\ z^{2} + \dfrac{5}{4}z& = \dfrac{14}{4} \\ (z + \dfrac{5}{8})^2 - (\dfrac{5}{8})^2 &=\dfrac{7}{2} \\ (z + \dfrac{5}{8})^2 &= (\dfrac{25}{64}) + \dfrac{7}{2} \\ (z + \dfrac{5}{8})^2& = \dfrac{249}{64} \end{align} Take the square root of both sides: \begin{align} z + \dfrac{5}{8} &= \pm \dfrac{\sqrt{249}}{8} \\ z &= -\dfrac{5}{8} \pm \dfrac{\sqrt{249}}{8}\\ z &=\dfrac{-5\pm \sqrt{249}}{8} \end{align} Hence solution set $=\left\{ \dfrac{-5\pm \sqrt{249}}{8}\right\}$

Solve the equation by completing square: $z^{2}=5 z-3$.

Solution.

\begin{align} z^{2} &= 5z - 3 \\ z^{2} - 5z &= -3 \\ (z - \dfrac{5}{2})^2 - \dfrac{25}{4} &= -3\\ (z - \dfrac{5}{2})^2& = \dfrac{13}{4} \end{align} Take the square root of both sides: \begin{align} z - \dfrac{5}{2}& = \pm \dfrac{\sqrt{13}}{2} \\ z &= \dfrac{5}{2} \pm \dfrac{\sqrt{13}}{2}\\ z &= \dfrac{5\pm \sqrt{13}}{2} \end{align} Hence solution set $\left\{\dfrac{5\pm \sqrt{13}}{2}\right\}$