Answer :
Answer:
The solutions to the quadratic equation are:
[tex]x=i+2,\:x=-i+2[/tex]
Step-by-step explanation:
Given the equation
- [tex]x^2-4x+5=0[/tex]
solving the equation
[tex]x^2-4x+5=0[/tex]
subtract 5 from both sides
[tex]x^2-4x+5-5=0-5[/tex]
[tex]x^2-4x=-5[/tex]
[tex]\mathrm{Add\:}\left(-2\right)^2\mathrm{\:to\:both\:sides}[/tex]
[tex]x^2-4x+\left(-2\right)^2=-5+\left(-2\right)^2[/tex]
[tex]\left(x-2\right)^2=-1[/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-2=\sqrt{-1}[/tex]
As the imaginary rule is given by
[tex]\:\sqrt{-1}=i[/tex]
so
[tex]x-2=i[/tex]
[tex]x=i+2[/tex]
solving
[tex]x-2=-\sqrt{-1}[/tex]
[tex]x-2=-i[/tex] ∵ [tex]\:\sqrt{-1}=i[/tex]
[tex]x=-i+2[/tex]
Therefore, the solutions to the quadratic equation are:
[tex]x=i+2,\:x=-i+2[/tex]