Question

How do you solve using the completing the square method `x^2 + 2x - 15 = 0`?

Answer 1

The solutions are:
`color(green)(x =3` or `color(green)(x = -5`

Explanation:

`x^2 + 2x - 15 =0`

`x^2 + 2x = 15`

To write the Left Hand Side as a Perfect Square, we add 1 to both sides:

`x^2 + 2x + 1 = 15 + 1`

`x^2 + 2 * x *1 + 1^2 = 16`

Using the Identity `color(blue)((a+b)^2 = a^2 + 2ab + b^2`, we get

`(x+1)^2 = 16`

`x + 1 = sqrt16` or `x + 1 = -sqrt16`

`color(green)(x = 4 - 1 =3` or `color(green)(x = 4 - 1 = -5`

Answer 2

`color(blue)(=> +x=-1+-4)`

`color(brown)(x= +3 " or "-5)`

Explanation:

The equation in standard form is:
`" "color(green)(y=ax^2+bx+c)`

The equation in vertex form is:
`" "color(blue)(y=a(x+b/(2a))^2+c-[(b/2)^2])`

`color(brown)("The "-[(b/2)^2]" corrects the error produced by" (b/(2a))^2`

They are different versions of the same thing!
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Solving your question")`

`a = 1" ; "b=2" ; "c=-15` giving

`y= (x+1)^2-15-1`

`color(blue)(y=(x+1)^2-16)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Set the vertex form equation as equal to zero

`=>0=(x+1)^2-16`

`=> (x+1)^2=+16`

Taking the square root of both sides

`=> +-(x+1)=+-sqrt(16)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Consider the case "-(x+1))`

`=> -x-1=+-4`

`=>-x=+1+-4`

Multiply both sides by (-1)

`color(blue)(=> +x=-1+-4)`

`color(brown)(x= +3 " or "-5)`

'....................................................................

`color(blue)("Consider the case "+(x+1))`

`=>x=-1+-4`

`color(brown)("Same as for "-(x+1))`
'...................................................