Question

How do you solve the equation `2(x+2)^2=72`?

Answer 1

`"answer "{x=-8" , "x=4}`

Explanation:

`2(x+2)^2=72`

`"divide both sides by 2"`

`(cancel(2)(x+2)^2)/cancel(2)=(72)/2`

`(x+2)^2=36`

`"take square root both sides"`

`(x+2)=+-6`

`"if "x+2=-6" , then " x=-8`

`"if "x+2=6 " , then " x=4`

Answer 2

`x = 4` or `-8`

Explanation:

`2(x+2)^2 = 72`

divide by `2`:

`(x+2)^2 = 36`

square root:

`x+2 = 6` or `-6`

subtract `2`:

`x = 4` or `-8`

Answer 3

`x={4,-8}`

Explanation:

First divide both sides by two to get `(x+2)^2=36`. Next take the square root of both sides to get `x+2=+-6`. Now split this into two equations for both the positive and negative six. These look like: `x+2=6` and `x+2=-6`. For both equations, subtract two from both sides which gives you `x=4` and `x=-8`. Hence `x={4,-8}`.

Answer 4

`x=-8" or " x=4`

Explanation:

Divide both sides by 2

`rArr2/2(x+2)^2=72/2`

`rArr(x+2)^2=36`

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

`sqrt((x+2)^2)=color(red)(+-)sqrt36`

`rArrx+2=color(red)(+-)6`

`• x+2=color(red)(6)`

subtract 2 from both sides.

`xcancel(+2)cancel(-2)=6-2`

`rArrx=4`

`• x+2=color(red)(-6)`

`rArrx=-8`

`color(blue)"as a check"`

Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.

`x=4to2(4+2)^2=2xx36=72to" true"`

`x=-8to2(-8+2)^2=2xx36=72to" true"`

`rArrx=-8" or " x=4" are the solutions"`