Question 779265
<pre>
There are two ways to do this problem.  The way you're probably
studying now is u-substitution:

       x-7&#8730;<span style="text-decoration: overline">x</span>+10 = 0

Let u = &#8730;<span style="text-decoration: overline">x</span>, then u² = x

       u²-7u+10 = 0

     (u-2)(u-5) = 0

  u-2 = 0;  u-5 = 0
    u = 2     u = 5

   &#8730;<span style="text-decoration: overline">x</span> = 2;     &#8730;<span style="text-decoration: overline">x</span> = 5 
 
(&#8730;<span style="text-decoration: overline">x</span>)² = 2²; (&#8730;<span style="text-decoration: overline">x</span>)² = 5²
    x = 4;      x = 25

We must always check radical
equations to find out if either
or both of those answers are 
legitimate solutions or are
only extraneous.

Checking x = 4:

       x-7&#8730;<span style="text-decoration: overline">x</span>+10 = 0
       4-7&#8730;<span style="text-decoration: overline">4</span>+10 = 0
       4-7·2+10 = 0
        4-14+10 = 0
              0 = 0

So 4 is a legitimate solution.

Checking x = 25:

       x-7&#8730;<span style="text-decoration: overline">x</span>+10 = 0
      25-7&#8730;<span style="text-decoration: overline">25</span>+10 = 0
      25-7·5+10 = 0
       25-35+10 = 0
              0 = 0 

So 25 is also a legitimate solution.

Edwin</pre>