Question 1010355
<pre>
{{{sqrt(x+7)+5 < x}}}

What's under the square root radical cannot be negative,
so we have a critical number when x+7=0 or x=7.

First we solve the equation of the boundary 
to find all other potential critical numbers besides -7.

{{{sqrt(x+7)+5 = x}}}

Isolate the square root term on the left:

{{{sqrt(x+7)=x-5}}}

Square both sides:

{{{x+7=x^2-10x+25}}}

{{{0=x^2-11x+18}}}

{{{0=(x-2)(x-9)}}}

From that we get critical values 2 and 9

We put all three critical values on a number line

-------o--------------------------o--------------------o------
-9 -8 -7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9 10 11

So we test the original inequality in each of the intervals 

Testing the inequality {{{(matrix(1,3,-infinity,",",-7))}}}
Choose test value -8 and substitute in the original inequality:

{{{sqrt(-8+7)+5 < -8}}}
{{{sqrt(-1)+5 < -8}}}

That's not a real number on the left, so we do not use that 
interval.

Testing the inequality {{{(matrix(1,3,-7,",",2))}}}
Choose test value 0 and substitute in the original inequality:

{{{sqrt(0+7)+5 < 0}}}
{{{sqrt(7)+5 < 0}}}

That's false, so we do not use that interval.

Testing the inequality {{{(matrix(1,3,2,",",9))}}}
Choose test value 3 and substitute in the original inequality:

{{{sqrt(3+7)+5 < 3}}}
{{{sqrt(10)+5 < 3}}}

That's false, so we do not use that interval.

Testing the inequality {{{(matrix(1,3,9,",",infinity))}}}
Choose test value 10 and substitute in the original inequality:

{{{sqrt(10+7)+5 < 10}}}
{{{sqrt(17)+5 < 10}}}

That's true, so we use that interval {{{(matrix(1,3,9,",",infinity))}}}

Now we test the critical values themselves:

Testing critical value -7

{{{sqrt(-7+7)+5 < -7}}}
{{{sqrt(0)+5<-7}}}

That's false.

Testing critical value 2

{{{sqrt(2+7)+5 < 2}}}
{{{sqrt(9)+5<2}}}

That's false.

Testing critical value 9

{{{sqrt(9+7)+5 < 9}}}
{{{sqrt(16)+5<9}}}
{{{4+5<9}}}

That's false. So the interval is open at 9

Answer: {{{(matrix(1,3,9,",",infinity))}}}

Edwin</pre>