Question

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

Answer 1

The solutions to the quadratic equation are:

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

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)=√(a),\:-√(a)`

solving

`x-2=√(-1)`

As the imaginary rule is given by

`\:√(-1)=i`

so

`x-2=i`

`x=i+2`

solving

`x-2=-√(-1)`

`x-2=-i` ∵
`\:√(-1)=i`

`x=-i+2`

Therefore, the solutions to the quadratic equation are:

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