Question

How do you solve `x^2+5x+8=0`?

Answer 1

`x = -5/2+-sqrt(7)/2i`

Explanation:

Pre-multiply by `4` (to avoid arithmetic with fractions), complete the square then use the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

with `a=(2x+5)` and `b=sqrt(7)i` as follows:

`0 = 4(x^2+5x+8)`

`=4x^2+20x+32`

`=(2x)^2+2(2x)(5)+32`

`=(2x+5)^2-25+32`

`=(2x+5)^2+7`

`=(2x+5)^2-(sqrt(7)i)^2`

`=((2x+5)-sqrt(7)i)((2x+5)+sqrt(7)i)`

`=(2x+5-sqrt(7)i)(2x+5+sqrt(7)i)`

So: `2x = -5+-sqrt(7)i`

So: `x = -5/2+-sqrt(7)/2i`

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Alternatively, use the quadratic formula:

The equation `x^2+5x+8 = 0` is of the form `ax^2+bx+c = 0` with `a=1`, `b=5` and `c=8`.

This has roots given by the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`(-5+-sqrt(5^2-(4*1*8)))/(2*1)`

`=(-5+-sqrt(25-32))/2`

`=(-5+-sqrt(-7))/2`

`=(-5+-sqrt(7)i)/2`

`=-5/2+-sqrt(7)/2i`