# 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: $$z = \dfrac{11 \pm \sqrt{7}i}{4}$$

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