Question

2Select the correct answer.What are the roots of this equation?x2 - 2x + 3 = 0OA. -2 + 12OB. 1 + 12OC. -1 2OD. 2 + V2Daca

Answer 1

Given the equation:

`x^2-2x+3=0`

Let's find the roots of the equation.

To find the roots of the equation, let's solve for x using the quadratic formula:

`x=(-b\pm√(b^2-4ac))/(4a)`

Use the standard equation to find the values of a, b, and c:

`\begin{gathered} ax^2+bx+c=0 \\ \\ x^2-2x+3=0 \end{gathered}`

a = 1

b = -2

c = 3

Input the values into the quadratic equation and solve for x:

`\begin{gathered} x=(-(-2)\pm√(-2^2-4(1)(3)))/(2(1)) \\ \\ x=(2\pm√(4-12))/(2) \\ \\ x=(2\pm√(-8))/(2) \end{gathered}`

Solving further:

`\begin{gathered} x=(2\pm√((8)(-1)))/(2) \\ \\ \text{ Where:} \\ √(-1)=i \\ \\ x=(2\pm i√(8))/(2) \\ \\ x=(2\pm i√((2)(4)))/(2) \\ \\ x=(2\pm i√((2)(2)^2))/(2) \end{gathered}`

Therefore, we have:

`\begin{gathered} x=(2\pm2i√(2))/(2) \\ \\ \text{ Divide all terms by 2:} \\ x=(2)/(2)\pm(2i√(2))/(2) \\ \\ x=1\pm i√(2) \end{gathered}`

The roots of the equation:

`x=1\pm i√(2)`

ANSWER:

`1\pm i√(2)`