Question

What are the possible values of x in 8x^2+4x=-1

Answer 1

`x1=-0.25(+)(0.25i)`

`x2=-0.25(-)(0.25i)`

Explanation:

we have

`8x^(2)+4x=-1`

Factor the leading coefficient

`8(x^(2)+0.5x)=-1`

Complete the square. Remember to balance the equation by adding the same constants to each side.

`8(x^(2)+0.5x+1/16)=-1+1/2`

`8(x^(2)+0.5x+1/16)=-1/2`

`(x^(2)+0.5x+1/16)=-1/16`

Rewrite as perfect squares

`(x+0.25)^(2)=-1/16`

remember that

`i=√(-1)`

`(x+0.25)=(+/-)(0.25i)`

`x=-0.25(+/-)(0.25i)`

`x1=-0.25(+)(0.25i)`

`x2=-0.25(-)(0.25i)`