Question

Solve`" " (-3x^2) + 4 = 0" "` by completing its square?

Answer 1

I am following the method even though some of the values are zero.

`-3(x+0)^2+4=0`

Explanation:

Insert the place keeper of `0x` as there aren't any.

Given:`" "(-3x^2)+0x+4=0`

Note that the brackets do not change anything so get rid of them

`-3x^2+0x+4=0`

Write as:

`-3(x^2+0/(-3)x)+k+4=0` where `k` is the correction factor

Take the square from `x^2` outside the brackets

`-3(x+0/(-3)x)^2+k+4=0`

Drop the `=0` for now due to the page width.

Halve the `0/(-3)`

`-3(x+0/(-3xx2)x)^2+k+4" "->" "color(red)(-3)(xcolor(blue)(-0/6))^2+k+4`

Set `" "color(red)((-3)color(blue)((-0/6)^2+k=0" "` neutralises the error.
`" "=>k=0` giving:

`-3(x+0x)^2+4=0`

Remove the `x` from `0x` giving:

`-3(x+0)^2+4=0`

`x_("vertex")=(-1)xx0=0`
`y_("vertex")=+4`

Vertex`->(x,y)=(0,4)`