Solve $\sqrt{x^2+10x+9}>|x+2|-5$

Question

Solve on $\mathbb R$: $\sqrt{x^2+10x+9}>|x+2|-5$

Can someone help me please?

I found that the interval where the inequality make sens is:

$D_f=]-\infty,-9[ \cup [-1,\infty[$.

then I have that, $\sqrt{x^2+10x+9}+5>|x+2|$

$\iff$ $$ \left\{ \begin{array}{c} \sqrt{x^2+10x+9}+5> x+2\\ -\sqrt{x^2+10x+9}-5<x+2 \end{array} \right. $$

But after many manipulation I dont have the right answer so I don't really understand why I don't have the right solution...

Answer 1

First of all we must have $x^2+10x+9\ge 0$

So $(x + 9)(x + 1) \ge 0$ so either $x+9 \ge 0$ and $x + 1 \ge 0$ and therefore $x \ge -9$ and $x \ge -1$ so $x \ge -1$ (as those two are redundant);

OR

$x+9 \le 0$ and $x+1 \le 0$ so so $x\le -9$ and $x \le -1$ so $x \le -9$ (as those two are redundant.

......

1) If $x \le -9$ then $x + 2 \le -7$ so $|x+2| = -x - 2 \ge 9 - 2 = 7$ so

$\sqrt{x^2 + 10x + 9} > -x - 7 \ge 2$

$x^2 + 10x + 9 > x^2 + 14x + 49 \ge 4$

$-40 > 4x$ and $x < -10$.

2) And if $x \ge -1$ then $x + 2 \ge 1$ so $|x + 2| = x+ 2$

so $\sqrt {x^2 + 10x + 9} > x -3$

2a) If $x \ge 3$ then $x^2 + 10x + 9 > x^2 - 6x +9$ so $16x > 0$. So $x \ge 3$

If $x < 3$ then $\sqrt {x^2 + 10x + 9} \ge 0> x -3$ and all we have is $x^2 + 10x + 9 \ge 0$ and $x \ge -1$

So putting that all together we have $x < -10$ or $x \ge -1$.

Answer 2

Splitting into two cases usually help when dealing with absolute values.

First case: $|x+2| = x+2$ or $x\geq -2.$ Now you can rearrange your inequality into: $$\sqrt{x^2+10x+9}>x-3,\,\, x\geq -1.$$

This is easy to solve and the remaining case would be: $$x<-2$$, which when intersected with your domain turns out to: $$x<-9.$$ In this region, you would solve the inequality: $$\sqrt{x^2+10x+9}>-7-x$$

After you solve both cases, you need to take the union of both solutions.

Answer 3

First we see that $x^2+10x+9 = (x+1)(x+9)\geq 0$ so $x\in (-\infty,-9]\cup[-1,\infty)$

From $\sqrt{x^2+10x+9}+5>|x+2|$ we get after squaring:

$$ x^2+10x+9 +10\sqrt{x^2+10x+9}+25>x^2+4x+4$$

or $$ 5\sqrt{x^2+10x+9}>-3x-15\;\;\;(*)$$

$\bullet$ If $x\leq -5$ we get (after squaring again) $$25x^2+250x+225>9x^2+90x+225$$ or

$$ x^2+10x> 0\implies x\notin [-10,0]$$

so in this case $\boxed{x<-10}$

$\bullet$ If $x> -5$ then $(*)$ is true for all $\boxed{x\geq -1}$

Answer 4

$\sqrt{x^2 + 10x + 9} \ge |x-2| - 5$

$x^2 + 10x + 9 \ge 0$ or the root will not be real.

$(x+1)(x+9) \ge 0\\ x\ge -1\text{ or } x\le -9$

$\sqrt{x^2 + 10x + 9} > x$ for all $x \ge -1$ and $|x+2| - 5 < x$ for all $x \ge -1$

$[-1,\infty)$ is in the solution set.

If $x<-9$

$|x-2| < 0$ or $|x+2| = -x-2$

$\sqrt{x^2 + 10x + 9} > -x -7$

Square both sides

$x^2 + 10x + 9 > x^2 + 14x + 49\\ 4x < -40\\ x<-10$

$(-\infty,-10)\cup [-1,\infty)$

Answer 5

The square root makes sense only for $x\le-9$ or $x\ge-1$.

For $x\le-9$ we have $|x+2|=-x-2$; for $x\ge-1$ we have $|x+2|=x+2$. So the inequality can be split into two cases: $$ \begin{cases} \sqrt{x^2+10x+9}\ge -x-7 \\[4px] x\le-9 \end{cases} \qquad\text{or}\qquad \begin{cases} \sqrt{x^2+10x+9}\ge x-3 \\[4px] x\ge-1 \end{cases} $$ In order to solve the first, note that $x\le-9$ implies $-x-7\ge0$; thus we can safely square: $$ \begin{cases} x^2+10x+9\ge x^2+14x+49 \\[4px] x\le-9 \end{cases} $$ that easily becomes $x\le-10$.

For the second, note that $-1\le x<3$ implies $x-3<0$, so the inequality $\sqrt{x^2+10x+9}\ge x-3$ is satisfied. Therefore, $[-1,3)$ is part of the solution set.

When $x\ge3$, we can square: $$ \begin{cases} \sqrt{x^2+10x+9}\ge x-3 \\[4px] x\ge3 \end{cases} $$ and after squaring $$ \begin{cases} x^2+10x+9\ge x^2-6x+9 \\[4px] x\ge3 \end{cases} $$ that becomes $x\ge 3$.

Thus the solution set is $$ (-\infty,-10]\cup[-1,3)\cup[3,\infty)= (-\infty,-10]\cup[-1,\infty) $$