Question

How do you solve `w^2-5=23`?

Answer 1

`w = 2sqrt7`

Explanation:

`w^2 - 5 = 23`

First, add `color(blue)5` to both sides of the equation:
`w^2 - 5 quadcolor(blue)(+quad5) = 23 quadcolor(blue)(+quad5)`

`w^2 = 28`

Now take the square root of both sides:
`sqrt(w^2) = sqrt28`

Now simplify to simplest radical form:
`w = sqrt(4*7)`

`w = sqrt4sqrt7`

`w = 2sqrt7`

Hope this helps!

Answer 2

`w=+-2sqrt7`

Explanation:

`"isolate "w^2" by adding 5 to both sides"`

`w^2=23+5=28`

`color(blue)"take the square root of both sides"`

`sqrt(w^2)=+-sqrt28larrcolor(blue)"note plus or minus"`

`w=+-sqrt(4xx7)=+-2sqrt7`