Question 130272
Let's start from where you are with this right now and work backwards.


You said that {{{sqrt(x^2-4x)=sqrt(8)}}}, and therefore {{{x-4x=4}}} or {{{x-4x=-4}}}


If {{{sqrt(8)=4}}} then {{{8 = 4 * 4}}}.  We know that isn't true, so the right side of your result is certainly incorrect.  {{{x-4x=-3x}}} and {{{(-3x)^2=9x}}}, and this is only true for two values of x (specifically 1 and -1).  Therefore, the right side of your result is also incorrect in general.


So let's go back one more step, the one before you took the square root of both sides.  {{{x^2-4x=8}}} is a good start.  But what you need to do now is 'complete the square'  That means that you are going to add a constant to both sides of the equation that makes that left side a perfect square.  The process for this is actually simpler than it sounds.


Step 1:  Divide both sides of your equation by the coefficient on the {{{x^2}}} term.  This is very easy for your problem because the {{{x^2}}} term coefficient is 1, and you don't have to change anything.


Step 2:  Take the coefficient on the {{{x}}} term and divide it by 2.  Your problem, the coefficient is -4, divided by 2 is -2.


Step 3:  Square the result of step 2.  For your problem, {{{(-2)^2=4}}}.


Step 4:  Add the result of step 3 to both sides of your equation.  Your problem:  {{{x^2-4x+4=12}}}


Step 5:  Now that the left side of the equation is a perfect square, you can take the square root of both sides.  First note that {{{x^2-4x+2=(x-2)^2}}} (You can use FOIL on {{{(x-2)(x-2)}}} to verify that fact)


{{{(x-2)^2=12}}}, so:


{{{x-2=sqrt(12)}}} or {{{x-2=-sqrt(12)}}}


{{{x=2+-sqrt(12)}}}


That is a representation of the solution set, in radical form (a radical is the square root sign).  However, there is one more simplification step that you can take.  Remember that {{{sqrt(ab)=sqrt(a)*sqrt(b)}}}.  12 = 4 * 3, so {{{sqrt(12)=sqrt(4)sqrt(3)=2*sqrt(3)}}}.  That means your final answer is:


{{{x=2+2*sqrt(3)}}} or {{{x=2-2*sqrt(3)}}}


Check the answer.  If we did the work correctly, we should be able to substitute either of these values for x into the original equation and have it be a true statement.


{{{x^2-4x-8=0}}}


{{{(2+2*sqrt(3))^2-4(2+2*sqrt(3))-8=0}}}


{{{(4+8*sqrt(3)+12)-(8+8*sqrt(3))-8=0}}}


{{{16+8*sqrt(3)-16-8*sqrt(3)=0}}}


{{{0=0}}}


So, the first answer works.  I'll leave it to you to prove the second one.  Or just trust me.