Question

Please help and explain

Answer 1

Answer: Option A

`x=(3+i)/(2)` or
`x=(3-i)/(2)`

Explanation:

Use the quadratic formula to find the zeros of the function.

For a function of the form

`ax ^ 2 + bx + c = 0`

The quadratic formula is:

`x=(-b\±√(b^2-4ac))/(2a)`

In this case the function is:

`2x^2-6x+5=0`

So

`a=2\\b=-6\\c=5`

Then using the quadratic formula we have that:

`x=(-(-6)\±√((-6)^2-4(2)(5)))/(2(2))`

`x=(6\±√(36-40))/(4)`

`x=(6\±√(-4))/(4)`

Remember that
`√(-1)=i`

`x=(6\±√(4)*√(-1))/(4)`

`x=(6\±√(4)i)/(4)`

`x=(6\±2i)/(4)`

`x=(3\±i)/(2)`

`x=(3+i)/(2)` or
`x=(3-i)/(2)`