# Question 3, Exercise 1.3

Solutions of Question 3 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 quadratic equation: $\dfrac{1}{3} z^{2}+2 z-16=0$.

Solution. Given \begin{align}&\dfrac{1}{3}z^{2}+2 z-16=0\\ \implies &z^{2} + 6z - 48 = 0 \end{align} Apply the quadratic formula: $$z = \dfrac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a},$$ where $$a = 1,\quad b = 6,\quad \text{and}\quad c = -48.$$ Then \begin{align} z& = \dfrac{{-6 \pm \sqrt{36-4(1)(-48)}}}{2 \cdot 1} \\ & = \dfrac{{-6 \pm \sqrt{228}}}{2 \cdot 1} \\ &= \dfrac{{-6 \pm 2\sqrt{57}}}{2} \\ &= -3 \pm \sqrt{57} \end{align} Hence Solution set $=\{ -3 \pm \sqrt{57} \}$.

Solve the quadratic equation: $z^{2}-\frac{1}{2} z+17=0$.

Solution.

Given $$z^{2} - \frac{1}{2}z + 17 = 0$$ Using the quadratic formula: $$z = \dfrac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}$$ Where
$$a = 1, \quad b = -\dfrac{1}{2}, \quad c = 17$$ Then \begin{align} z &= \dfrac{{-\left(-\dfrac{1}{2}\right) \pm \sqrt{{\left(-\dfrac{1}{2}\right)^2 - 4 \cdot 1 \cdot 17}}}}{2 \cdot 1} \\ &= \dfrac{{\dfrac{1}{2} \pm \sqrt{{\dfrac{1}{4} - 68}}}}{2} \\ &= \dfrac{{\dfrac{1}{2} \pm \sqrt{{\dfrac{1 - 272}{4}}}}}{2} \\ &= \dfrac{{\dfrac{1}{2} \pm \sqrt{{\dfrac{-271}{4}}}}}{2} \\ &= \dfrac{{\dfrac{1}{2} \pm \dfrac{\sqrt{-271}}{2}}}{2} \\ &= \dfrac{1 \pm \sqrt{271}i}{4} \end{align}

Therefore, the solution set is: $\left\{\dfrac{1 \pm \sqrt{271}i}{4}\right\}$

Solve the quadratic equation: $z^{2}-6 z+25=0$.

Solution.

Given $$z^{2} - 6z + 25 = 0$$ Using the quadratic formula: $$z = \dfrac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}$$ Where $$a = 1, \quad b = -6, \quad c = 25$$ Then \begin{align} z &= \dfrac{{-(-6) \pm \sqrt{{(-6)^2 - 4 \cdot 1 \cdot 25}}}}{2 \cdot 1} \\ &= \dfrac{{6 \pm \sqrt{{36 - 100}}}}{2} \\ &= \dfrac{{6 \pm \sqrt{{-64}}}}{2} \\ &= \dfrac{{6 \pm 8i}}{2} \\ &= 3 \pm 4i \end{align}

Therefore, the solution set is: $\{3 \pm 4i\}$

Solve the quadratic equation: $z^{2}-9 z+11=0$.

Solution.

Given $$z^{2} - 9z + 11 = 0$$ Using the quadratic formula: $$z = \dfrac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}$$

Where $$a = 1, \quad b = -9, \quad c = 11$$ Then \begin{align} z &= \dfrac{{-(-9) \pm \sqrt{{(-9)^2 - 4 \cdot 1 \cdot 11}}}}{2 \cdot 1} \\ &= \dfrac{{9 \pm \sqrt{{81 - 44}}}}{2} \\ &= \dfrac{{9 \pm \sqrt{37}}}{2} \end{align}

Therefore, the solution set $=\left\{ \dfrac{9 \pm \sqrt{37}}{2}\right\}$