Question 1199266
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Solve for x this equation  x^2 + 3x + sqrt{x^2 + 3x} = 6.
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            The method to solve is introducing new variable.



<pre>
(1)  Notice that the expression under the square root must be non-negative

        x^2 + 3x >= 0,  or, equivalently,  x*(x+3) >=0,

    which means that either x >= 0  or x <= -3.

    Thus the domain is  (-oo,-3] U [0,oo).


    Also, notice that we consider positive values of  {{{sqrt(x^2+3x)}}}, only, so y >= 0.



(2)  Introduce new variable  y = {{{sqrt(x^2 + 3x)}}}.  

     Then  y^2 = x^2 + 3x,  and the given equation takes the form

         y^2 + y = 6,   or, equivalently,

         y^2 + y - 6 = 0


     Factor left sides

         (y+3)*(y-2) = 0

     The roots are  y= -3  ans  y= 2.


     Since we consider only positive values of y in this problem, 
     we analyse below the case y= 2 ONLY.



(3)  If y= 2, then

         {{{sqrt(x^2 + 3x)}}} = 2.


     Square both sides, simplify and find x

         x^2 + 3x = 4

         x^2 + 3x - 4 = 0

         (x+4)*(x-1) = 0

     Two roots are  x= -4,  x= 1.         
</pre>

Solved, and the obtained answer coincides with that in your textbook.


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Memorize the method of the solution: it is introducing new variable 
and accurate analysis of the domains of the variables x and y.