<PRE><B>Question 1127948</B>
&lt;br&gt;
{{{sqrt(3x+1)-x+3 = 0}}}&lt;br&gt;
Isolate the radical&lt;br&gt;
{{{sqrt(3x+1) = x-3}}}&lt;br&gt;
Square both sides to get rid of the radical&lt;br&gt;
{{{3x+1 = x^2-6x+9}}}&lt;br&gt;
Solve the quadratic equation by factoring&lt;br&gt;
{{{x^2-9x+8 = 0}}}
{{{(x-8)(x-1) = 0}}}&lt;br&gt;
The two possible solutions are x=8 and x=1.&lt;br&gt;
Because we squared both sides of the equation at one point, there might be extraneous roots.  So see if each of the possible roots satisfies the original equation.&lt;br&gt;
x=8:  {{{sqrt(25)-8+3 = 5-8+3 = 0}}}  YES&lt;br&gt;
x=1:  {{{sqrt(4)-1+3 = 2-1+3 &lt;&gt;0}}}  NO&lt;br&gt;
Note if we had been able to use -2 as the square root of 4, the equation would have been satisfied: -2-1+3 = 0.  This will usually (maybe always, I&#39;m not sure) be the case when you solve a radical equation like this and one of the roots is extraneous.&lt;br&gt;
ANSWER: Choice A: one solution, x=8; x=1 is extraneous</PRE>