Question

Question #2236d

Answer 1

There are a few different ways to approach it, but I've shown one below.

In this case the solution is `x=-5.83` or `x=-0.17`

Explanation:

Equations can be solved in different ways, as long as whatever we do to one side we do to the other. Here's one way to solve this one:

`2(x+3)^2=16`

Divide both sides by 2:

`(x+3)^2=8`

Expand and simplify the brackets:

`(x+3)(x+3)=8`

`x^2+3x+3x+9=8`

`x^2+6x+9=8`

Subtract 8 from both sides:

`x^2+6x+1=0`

This is a standard form for quadratic equations, `ax^2+bx+c=0`, and allows us to use the quadratic formula to solve the equation:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`=(-6+-sqrt(36-4xx1xx1))/(2)`

`=(-6+-sqrt(32))/(2)`

That means `x=-5.83` or `x=-0.17`

Answer 2

`x=-3+-2sqrt2`

Explanation:

`"isolate "(x+3)^2" by dividing both sides of the equation by 2"`

`cancel(2)/cancel(2)(x+3)^2=16/2`

`rArr(x+3)^2=8`

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

`sqrt((x+3)^2)=+-sqrt8larrcolor(blue)" note plus or minus"`

`rArrx+3=+-sqrt8=+-sqrt4xxsqrt2=+-2sqrt2`

`"subtract 3 from both sides"`

`xcancel(+3)cancel(-3)=-3+-2sqrt2`

`rArrx=-3-2sqrt2" or "x=-3+2sqrt2`

`rArrx~~ -5.828" or "x~~ -0.172" to 3 dec. places"`