Question

What are the solutions to x^2-2x+5=0

Answer 1

x = 1 ± 2i

Explanation:

x² - 2x + 5 = 0 ( add 5 to both sides )

x² - 2x = - 5

Using the method of completing the square

add ( half the coefficient of the x- term)² to both sides

x² + 2(- 1)x + 1 = - 5 + 1

(x - 1)² = - 4 ( take square root of both sides )

x - 1 = ±
`√(-4)` = ± 2i ( add 1 to both sides )

x = 1 ± 2i ( that is the solutions are complex )

Answer 2

`x=1\pm2i`

or

`x=1+2i`

`x=1-2i`

Explanation:

All equations of the form
`ax^2+bx+c=0` can be solved using the quadratic formula:
`x = (-b \pm √(b^2-4ac))/(2a)`

Quadratic formula gives two solutions, one when ± is addition add one when it is subtraction.

`x^2-2x+5=0`

Equation is in standard form:
`ax^2+bx+c=0`

Substitute 1 from a,
`-2` for b, and
`5` for C in the quadratic formula,
`(-b\pm √(b^2-4ac))/(2a)`

`x=(-\left(-2\right)\pm √(\left(-2\right)^2-4* \:1* \:5))/(2)`

Multiply -4 × 5:

`x=(-\left(-2\right)\pm √(\left(-2\right)^2-4\cdot \:1\cdot \:5))/(2)`

Add 4 + -20:

`x=(-\left(-2\right)\pm \:4i)/(2)`

* the opposite of -2 is 2:

`x=(2\pm 4i)/(2)`

Now, solve equation:

`x=(2+4i)/(2)`

Divide 2 + 4i by 2:

`x=1-2i`

_____________________________