<PRE><B>Question 540815</B>
{{{x=2sqrt(x)+3}}}&lt;P&gt;
{{{x-3=2sqrt(x)}}}&lt;P&gt;
Square both sides&lt;P&gt;
{{{(x-3)^2 = 4x}}}&lt;P&gt;
{{{x^2 - 6x + 9 = 4x}}}&lt;P&gt;
{{{x^2 - 10x + 9 = 0}}}&lt;P&gt;
{{{(x-9)(x-1)=0}}}&lt;P&gt;
x-9=0 and x-1=0&lt;P&gt;
x=9 and x=1&lt;P&gt;
Try them in the original equation.  9 works.  1 works because recall that {{{sqrt(x)}}} can be positive or negative.  If it&#39;s negative (-1 * -1 = 1) then the original equation is:
1 = -2+3 which is correct.
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