Question

What are the solutions of x2-4x+5=0 please help!!

Answer 1

Answer:

The solutions to the quadratic equation are:

$x=i+2,\:x=-i+2$

Step-by-step explanation:

Given the equation

• $x^2-4x+5=0$

solving the equation

$x^2-4x+5=0$

subtract 5 from both sides

$x^2-4x+5-5=0-5$

$x^2-4x=-5$

$\mathrm{Add\:}\left(-2\right)^2\mathrm{\:to\:both\:sides}$

$x^2-4x+\left(-2\right)^2=-5+\left(-2\right)^2$

$\left(x-2\right)^2=-1$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-2=\sqrt{-1}$

As the imaginary rule is given by

$\:\sqrt{-1}=i$

so

$x-2=i$

$x=i+2$

solving

$x-2=-\sqrt{-1}$

$x-2=-i$      ∵ $\:\sqrt{-1}=i$

$x=-i+2$

Therefore, the solutions to the quadratic equation are:

$x=i+2,\:x=-i+2$