Answer:
The solutions to the quadratic function are:
[tex]x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}[/tex]
Step-by-step explanation:
Given the function
[tex]f\left(x\right)\:=\:4x^2\:-\:4x\:+\:3[/tex]
Let us determine all possible solutions for f(x) = 0
[tex]0=4x^2-4x+3[/tex]
switch both sides
[tex]4x^2-4x+3=0[/tex]
subtract 3 from both sides
[tex]4x^2-4x+3-3=0-3[/tex]
simplify
[tex]4x^2-4x=-3[/tex]
Divide both sides by 4
[tex]\frac{4x^2-4x}{4}=\frac{-3}{4}[/tex]
[tex]x^2-x=-\frac{3}{4}[/tex]
Add (-1/2)² to both sides
[tex]x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{3}{4}+\left(-\frac{1}{2}\right)^2[/tex]
[tex]x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{1}{2}[/tex]
[tex]\left(x-\frac{1}{2}\right)^2=-\frac{1}{2}[/tex]
[tex]\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}[/tex]
solving
[tex]x-\frac{1}{2}=\sqrt{-\frac{1}{2}}[/tex]
[tex]x-\frac{1}{2}=\sqrt{-1}\sqrt{\frac{1}{2}}[/tex] ∵ [tex]\sqrt{-\frac{1}{2}}=\sqrt{-1}\sqrt{\frac{1}{2}}[/tex]
as
[tex]\sqrt{-1}=i[/tex]
so
[tex]x-\frac{1}{2}=i\sqrt{\frac{1}{2}}[/tex]
Add 1/2 to both sides
[tex]x-\frac{1}{2}+\frac{1}{2}=i\sqrt{\frac{1}{2}}+\frac{1}{2}[/tex]
[tex]x=i\sqrt{\frac{1}{2}}+\frac{1}{2}[/tex]
also solving
[tex]x-\frac{1}{2}=-\sqrt{-\frac{1}{2}}[/tex]
[tex]x-\frac{1}{2}=-i\sqrt{\frac{1}{2}}[/tex]
Add 1/2 to both sides
[tex]x-\frac{1}{2}+\frac{1}{2}=-i\sqrt{\frac{1}{2}}+\frac{1}{2}[/tex]
[tex]x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}[/tex]
Therefore, the solutions to the quadratic function are:
[tex]x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}[/tex]
