Question

How do you solve `w ^ { 2} - 6w + 9= 13`?

Answer 1

`w = 3+-sqrt(13)`

Explanation:

Given:

`w^2-6w+9 = 13`

Note that the left hand side is already a perfect square trinomial, so we have:

`(w-3)^2 = 13`

Taking the square root of both sides, allowing for both possible square roots, we have:

`w-3 = +-sqrt(13)`

Then add `3` to both sides to get:

`w = 3+-sqrt(13)`

Answer 2

`w=3+-sqrt13`

Explanation:

`w^2-6w+9=13`
Rearrange
`w^2-6w-4=0`
Either use the formula `w=(-b+-sqrt(b^2-4ac))/(2a)`
Where `aw^2+bw+c=0`
`w=(6+-sqrt(36-4*1*-4))/2`
`w=(6+sqrt(36+16))/2` or `w=(6-sqrt(36+16))/2`
`w=(6+sqrt52)/2=3+sqrt13`
Or
`w=3-sqrt13`

The other way is completing the square ( which is how the formula is derived)
`w^2-6w-4=0`
`(w-3)^2=w^2-6w+9`
So `w^2-6w-4` can be written `(w-3)^2-13=0`
`(w-3)^2=13`
Take the square root of both side
`w-3=+-13`
`w=3+-13`