Question

How do you solve by completing the square: `2x^2 + 8x + 1 = 0`?

Answer 1

`2x^2 + 8x +1 = 0`

Reduce the coefficient of `x^2` to `1` by dividing all terms by `2`
#x^2+4x +1/2 = 0

Remove the constant `1/2` from the left-side by subtracting `1/2` from both sides of the equation.
`x^2+4x = -1/2`

To "complete the square" we are looking for a value `a`
`(x+a)^2 = x^2 +2ax +a^2`

From our equation we know that `2ax = 4x`
`rarr a=2`
and `a^2 = 4`

Add `a^2` (`4`) to both sides of the equation to "complete the square"
`x^2 + 4x +4 = 4-1/2`
or
`(x+2)^2 = 7/2`

Take the square root of both sides
`x+2` = +-sqrt(7/2)#

Therefore
`x = -2 -sqrt(7/2) = - (sqrt(2)+sqrt(7))`
or
`x = -2 + sqrt(7/2) = sqrt(7) - sqrt(2)`