How to find the solution of $\sqrt{x^2+12x+35} \geq x-10$? I want to ask about radical inequality problem.
Here's the question:
Find the solution sets for $\sqrt{x^2+12x+35}\geq x-10$
My attempts to tackle this problem is like this:
Firstly I try to squaring this inequality such that:
$x^2+12x+35\geq (x-10)^2$
$x^2+12x+35\geq x^2-20x+100$
$x=\frac{65}{32}$
And then I apply the condition for the form under the radical, such that:
$x^2+12x+35\geq 0$ and also for the right hand, such that: $x-10\geq 0$
I solve for both of them and got $x\leq -7$ and $x\geq -5$ for the first one and $x>=10$ for the right hand.
Then, I combine all of the solutions to get whole solutions $x>=10$
From my number line, I can conclude that the solution should be $x\geq 10$
But, when I'm trying to check it in wolfram, I got the solution should be $x\leq 7$ or $x\geq -5$
Please somebody explain to me about this difference, so I can get the right answer from this problem. Thanks anyway.
 A: First you have to sate the constraint on your inequality . This inequality make senses only when term under square root is nonnegative . 
i.e $$x^2+12x+35\ge 0\Longleftrightarrow (x+5)(x+7)\ge 0\Longleftrightarrow x\in(-\infty, -7]\cup [-5, \infty)$$
Adding to this it becomes obvious when the left hand side is non-positif. ie if 
$$ x-10\le 0 \Longleftrightarrow x\in (-\infty ,10]$$
So, if $x\in (-\infty ,10]\cap((-\infty, -7]\cup [-5, \infty)) =(-\infty, -7]\cup [-5,10]$ we have 
 $\sqrt{x^2+12x+35} >= x-10$
Now if $x\in (-\infty, -7]\cup [-5, \infty) $ and $x\in  [10, \infty)$ which means $x\in [10, \infty) $ and $x-10\ge0$ you can proceed as you did i.e 
For $x\in [10, \infty)$ we have, 
$$\sqrt{x^2+12x+35} >= x-10 \Longleftrightarrow x^2+12x+35\ge x^2-20x+100\Longleftrightarrow 32x \ge 65$$
Then, $x\ge 65/32 =2.03125$ and  $x\in [10, \infty)$ this implies that $x\in [10, \infty)$

finally, $\sqrt{x^2+12x+35} >= x-10 \Longleftrightarrow x\in [10, \infty)\cup (-\infty, -7]\cup [-5,10] =(-\infty, -7]\cup [-5,\infty)$

A: We can write your equations as 
$$\sqrt{(x+6)^2-1}\ge x-10$$
For us to be able to write $$\sqrt{(x+6)^2-1}$$
It must be that $$ \bbox[5px,border:2px solid red]{x\in \Big(-\infty ,-7\Big]\cup \Big[-5,\infty \Big)}$$
$$$$$$$$
For $$ -5\le x\le 10$$
$$\sqrt{(x+6)^2-1}\ge 0\ge x-10$$
Which is true.
Hence,
$$ \bbox[5px,border:2px solid red]{x\in \Big[-5,10\Big]}$$
$$$$$$$$
For $x\gt 10$
$$\sqrt{(x+6)^2-1}\ge x-10$$
$$(x+6)^2-1\ge (x-10)^2$$
$$(x+6)^2-(x-10)^2 \ge 1$$
$$((x+6)+(x-10))((x+6)-(x-10))\ge 1$$
$$(2x-4)16\ge 1$$
$$32x-64\ge 1$$
$$32x\ge 65$$
$$ x\ge \frac{65}{32}$$
But
$$ \bbox[5px,border:2px solid red]{x\in \Big[10,\infty \Big)}$$

Considering all conditions,
  $$x\in \Big(-\infty,-7\Big]\cup \Big[-5,\infty\Big)$$

A: Hints:


*

*First, you have to determine the domain of validity of the inequation, i.e. for which values  of $x$ the radicand is non-negative. As 
$$x^2+12x+35=(x+5)(x+7),$$
this domain is $\;(-\infty,-7]\cup[-5,+\infty)$.

*Second, you must remember that, on the domain of validity,
$$\sqrt A\ge B\iff A\ge B^2\quad\textbf{or}\quad B\le 0.$$
A: Let $x^2+12x+35=y^2$, where $y\geq0$.
Thus, $x=-6-\sqrt{y^2+1}$ or $x=-6+\sqrt{y^2+1}$ and we need to solve
$$y+16\geq\sqrt{y^2+1}$$ and we need to solve
$$y+16\geq-\sqrt{y^2+1}.$$
But both inequalities are obviously true for all $y\geq0$, 
which says that the starting inequality is equivalent to
$$x^2+12x+35\geq0,$$ which gives the answer:
$$(-\infty,-7]\cup[-5,+\infty).$$
Done!
