====== 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. 

====Question 2(i)====
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=$\{3 \pm \sqrt{7}\}$.


====Question 2(ii)====
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$=\{-5 \pm \sqrt{29}\}$


====Question 2(iii)====
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 $=\left\{ \dfrac{-5\pm \sqrt{249}}{8}\right\}$

====Question 2(iv)====
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 $\left\{\dfrac{5\pm \sqrt{13}}{2}\right\}$



====Go to ====
<text align="left"><btn type="primary">[[math-11-nbf:sol:unit01:ex1-3-p1|< Question 1]]</btn></text>
<text align="right"><btn type="success">[[math-11-nbf:sol:unit01:ex1-3-p3|Question 3 >]]</btn></text>