Question

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

Answer 1

Answer:

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

$x2=-0.25(-)(0.25i)$

Step-by-step 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

$i=\sqrt{-1}$

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

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

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

$x2=-0.25(-)(0.25i)$