Question

How do you solve `4a ^ { 2} + 4a + 5= 0`?

Answer 1

`a=-1/2+-i`

Explanation:

Given:

`4a^2+4a+5=0`

We can complete the square and use the difference of squares identity, but the solutions are non-real Complex ones.

The difference of squares identity can be written:

`A^2-B^2=(A-B)(A+B)`

Use this with `A=(2a+1)` and `B=2i` (where `i^2=-1`) as follows:

`0 = 4a^2+4a+5`

`color(white)(0) = 4a^2+4a+1+4`

`color(white)(0) = (2a)^2+2(2a)+1^2+2^2`

`color(white)(0) = (2a+1)^2-(2i)^2`

`color(white)(0) = ((2a+1)-2i)((2a+1)+2i)`

`color(white)(0) = (2a+1-2i)(2a+1+2i)`

Hence:

`2a=-1+-2i`

Dividing both sides by `2` we get:

`a=-1/2+-i`