Question

How do you find all the real and complex roots of `x^2 + 8x + 25 = 0`?

Answer 1

`x=-4+-3i`

Explanation:

Complete the square.

`(x^2+8x+16)+25-16=0`

`(x+4)^2+9=0`

`(x+4)^2=-9`

`sqrt((x+4)^2)=sqrt(-9)`

`x+4=+-3i`

`x=-4+-3i`

Answer 2

Roots (both complex) are at `x=-4+3i` and `x=-4-3i`

Explanation:

The quadratic formula tells us that for a quadratic equation in the form:
`color(white)("XXX")ax^2+bx+c=0`
the roots are given by:
`color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)`

For the given example: `x^2+8x+25=0`
`color(white)("XXX")a=1, b=8, and c=25`

So the roots are:
`color(white)("XXX")x=(-8+-sqrt(8^2-4(1)(25)))/(2(1))`

`color(white)("XXXX")=(-8+-sqrt(-36))/2`

`color(white)("XXXX")=(-8+-sqrt(36)*sqrt(-1))/2`

`color(white)("XXXX")=-4+-3i`