Answer:
C.
[tex]x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i[/tex] and [tex]x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i[/tex]
Step-by-step explanation:
You have the quadratic function [tex]2x^2-x+1=0[/tex] to find the solutions for this equation we are going to use Bhaskara's Formula.
For the quadratic functions [tex]ax^2+bx+c=0[/tex] with [tex]a\neq 0[/tex] the Bhaskara's Formula is:
[tex]x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}[/tex]
[tex]x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}[/tex]
It usually has two solutions.
Then we have [tex]2x^2-x+1=0[/tex] where a=2, b=-1 and c=1. Applying the formula:
[tex]x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}[/tex]
Observation: [tex]\sqrt{-1}=i[/tex]
[tex]x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i[/tex]
And,
[tex]x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i[/tex]
Then the correct answer is option C.
[tex]x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i[/tex] and [tex]x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i[/tex]
