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

2 Answers
May 24, 2018

Answer:

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#

May 24, 2018

Answer:

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#