Question

How do you solve `\sqrt { ( x - 2) ( x + 4) } - x - 2= 0`?

Answer 1

No solutions

Explanation:

writing `sqrt((x-2)(x+4))=x+2`
by squaring we get
`x^2+2x-8=x^2+4x+4`
or
`-12=2x` so `x=-6`
This fulfilles not our equation since
`sqrt((-8)(-2))ne -4`

Answer 2

Move values that are not under the radical to the other side and square both sides. After Rearranging, you will find `x=-6`

Explanation:

First, move the values that aren't under the radical to the Right Hand Side (RHS):

`sqrt((x-2)(x+4))cancel(-x-2)color(red)(cancel(+x+2))=0color(red)(+x+2)`

`sqrt((x-2)(x+4))=x+2`

Next, we'll FOIL the two expressions that are under the radical:

`sqrt(x^2+2x-8)=x+2`

Next, we'll square both sides to get rid of the radical:

`(sqrt(x^2+2x-8))^2=(x+2)^2`

`x^2+2x-8=x^2+4x+4`

We can eliminate the `x^2` term since its the same for both sides. We will also begin to rearrange the equation so all of the `x`-terms are on one side and all of the constants are on the other:

`cancel(2x)-8color(blue)(cancel(-2x))color(red)(-4)=4xcancel(+4)color(blue)(-2x)color(red)(cancel(-4))`

`-8color(red)(-4)=4xcolor(blue)(-2x)`

`-12=2x`

Finally, divide both sides by `x`'s coefficient:

`-12/color(red)(2)=(cancel(2)x)/color(red)(cancel(2))`

`color(green)(x=-6)`

Let's apply this back in the (somewhat) original equation:

`sqrt((x-2)(x+4))=x+2`

`sqrt((color(green)((-6))-2)(color(green)((-6))+4))=color(green)((-6))+2`

`sqrt((-8)(-2))=-4`

`sqrt(16)=-4`

Now, here's a tricky part. The square root of a number can be either positive or negative, since any real number squared is ALWAYS positive.

`+-4=-4`

`-4=-4`