Question 14 Marks
Show the following quadratic equation:
$\text{x}^2-(3\sqrt{2}-2\text{i})\text{x}-\sqrt{2}\text{ i}=0$
$\text{x}^2-(3\sqrt{2}-2\text{i})\text{x}-\sqrt{2}\text{ i}=0$
Answer
View full question & answer→We will apply discriminate rule on $ax^2 + bx + c = 0$
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
Now,
$\text{x}^2-(3\sqrt{2}-2\text{i})\text{x}-\sqrt{2}\text{ i}=0$
$\text{x}=\frac{(3\sqrt{2}-2\text{i})\pm\sqrt{[-(3\sqrt{2}-2\text{i})]^2-4(1)(-\sqrt{2})\text{i}}}{2(1)}$
$=\frac{(3\sqrt{2}-2\text{i})\pm\sqrt{(3\sqrt{2})-2\text{i})^2+4\sqrt{2}\text{i}}}{2}$
$=\frac{3\sqrt{2}-2\text{i}}{2}\pm\frac{4-\sqrt{2}\text{i}}{2}$
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
Now,
$\text{x}^2-(3\sqrt{2}-2\text{i})\text{x}-\sqrt{2}\text{ i}=0$
$\text{x}=\frac{(3\sqrt{2}-2\text{i})\pm\sqrt{[-(3\sqrt{2}-2\text{i})]^2-4(1)(-\sqrt{2})\text{i}}}{2(1)}$
$=\frac{(3\sqrt{2}-2\text{i})\pm\sqrt{(3\sqrt{2})-2\text{i})^2+4\sqrt{2}\text{i}}}{2}$
$=\frac{3\sqrt{2}-2\text{i}}{2}\pm\frac{4-\sqrt{2}\text{i}}{2}$