Answer:
[tex]x= \dfrac{2 \pm i\sqrt{17}}{3}[/tex]
Step-by-step explanation:
A complex number is the sum of a real number and an imaginary number.
An imaginary number is a non-zero real number multiplied by the imaginary unit [tex]i[/tex].
[tex]i=\sqrt{-1} \implies i^2=-1[/tex]
For example:
- 5[tex]i[/tex] is an imaginary number (and its square is -25)
- 5 + 5[tex]i[/tex] is a complex number
Quadratic Formula
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
Given quadratic equation:
[tex]-3x^2+4x-7=0[/tex]
Therefore:
[tex]a=-3, \quad b=4, \quad c=-7[/tex]
Inputting these values into the quadratic formula:
[tex]\begin{aligned}\implies x & =\dfrac{-4 \pm \sqrt{4^2-4(-3)(-7)}}{2(-3)}\\\\ & = \dfrac{-4 \pm \sqrt{-68}}{-6}\\\\ & = \dfrac{-4 \pm \sqrt{4 \cdot -17}}{-6}\\\\ & = \dfrac{-4 \pm \sqrt{4}\sqrt{-17}}{-6}\\\\ & = \dfrac{-4 \pm 2\sqrt{-17}}{-6}\\\\ & = \dfrac{2 \pm \sqrt{-17}}{3}\\\\ & = \dfrac{2 \pm \sqrt{-1 \cdot 17}}{3}\\\\ & = \dfrac{2 \pm \sqrt{-1}\sqrt{17}}{3}\\\\ & = \dfrac{2 \pm i\sqrt{17}}{3}\end{aligned}[/tex]
Therefore, the solutions to the given quadratic equation are complex numbers as they include the imaginary number [tex]i\sqrt{17}[/tex].
