====== Question 7, Review Exercise ======
Solutions of Question 7 of Review Exercise 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 7 =====
Solve by completing square method $2 z^{2}-11 z+16=0$.
 
** Solution. **

\begin{align*}
&2 z^{2}-11 z+16=0\\
\implies&z^2 - \dfrac{11}{2}z + 8 = 0\\
\implies& z^2 - \dfrac{11}{2}z = -8\\
\implies& z^2 - 2z\dfrac{11}{4}z + \dfrac{121}{16} = -8 + \dfrac{121}{16}\\
\implies&\left(z-\dfrac{11}{4}\right)^2 =  -\dfrac{128}{16} + \dfrac{121}{16}\\
\implies&\left(z - \dfrac{11}{4}\right)^2 = -\dfrac{7}{16} 
\end{align*}
Take the square root of both sides:
\begin{align*}
&z - \dfrac{11}{4} = \pm \sqrt{-\dfrac{7}{16}}\\
\implies& z = \dfrac{11}{4} \pm \dfrac{\sqrt{7}i}{4}\\
\implies& z = \dfrac{11 \pm \sqrt{7}i}{4}
\end{align*}
So the solutions are:
$$z = \dfrac{11 \pm \sqrt{7}i}{4}$$
GOOD

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