Question 1094302
<pre>
Solving by substitution with u
{{{ x^2-3x-sqrt(x^2-3x)=2 }}}

{{{ u = sqrt(x^2-3x) }}}
{{{ u^2 = sqrt(x^2-3x) = x^2-3x }}}
creates quadratic = {{{ u^2-u = 2 }}}
{{{ u^2-u-2=0 }}}
{{{ (u-2)(u+1) }}}
u = 2,-1
going back and using those two outputs in {{{ sqrt(x^2-3x) }}}
{{{ sqrt((2)^2-3(2)) }}} and {{{ sqrt((-1)^2-3(-1)) }}}
{{{ sqrt(4-6) }}} and {{{ sqrt(1+3) }}}
Finally, I get {{{ sqrt(-2) }}} and {{{ sqrt(4) }}}

But -2 can't be in a radical and would 4 turn into 2,-2 or just 2? 
******************************************************************<font color = blue><font size = 2><font face = tahoma><b>

u = 2,-1 <=== This is okay!
"going back and using those two outputs in {{{ sqrt(x^2-3x) }}}    {{{ sqrt((2)^2-3(2)) }}} and {{{ sqrt((-1)^2-3(-1)) }}} <=== Here's where I guess 
                                                                                                                you got confused, and substituted 2 and - 1 for x in {{{sqrt(x^2-3x)}}}

But, u = 2, as you mentioned above, NOT x = 2. And, because you'd substituted u for {{{sqrt(x^2 - 3x)}}} earlier, at this juncture, you
need to BACK-SUBSTITUTE the value of u to get:
                      {{{2 = sqrt(x^2 - 3x)}}} 
                   {{{2^2 = (sqrt(x^2 - 3x))^2}}} ---- Squaring both sides
                    {{{4 = x^2 - 3x}}}
     {{{x^2 - 3x - 4 = 0}}}
(x - 4)(x + 1) = 0
  x - 4 = 0          OR        x + 1 = 0
       x = 4           OR              x = - 1 

Now, you have 2 values for x that you can CHECK to ensure that they're VALID and NOT EXTRANEOUS. 

Also, u = - 1, as you mentioned above. And, because you'd substituted u for {{{sqrt(x^2 - 3x)}}} earlier, at this juncture, you need 
to BACK-SUBSTITUTE the value of u to get: {{{- 1 = sqrt(x^2 - 3x)}}} 
Seeing that the square root of ANY expression is positive (> 0), it's obvious that u = - 1 is an EXTRANEOUS value. As such,
x = 4, or x = - 1 (see above). </font></font></font></b></pre>