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.
Question 3(i)
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 $=\{ -3 \pm \sqrt{57} \}$.
Question 3(ii)
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\} $
Question 3(iii)
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: $\{3 \pm 4i\}$
Question 3(iv)
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 $=\left\{ \dfrac{9 \pm \sqrt{37}}{2}\right\}$
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