# Solve $x+3<\sqrt{x+33}$ : Solution Verification

First of all the part inside the square root should be greater than or equal to $$0$$. So,
$$x+33 \ge0$$
$$\Rightarrow x \ge -33$$

Now I solved the inequality,
$$x+3<\sqrt{x+33}$$
$$\Rightarrow(x+3)^2<{x+33}$$
$$\Rightarrow x^2+9+6x<{x+33}$$
$$\Rightarrow x^2+5x-24 < 0$$
$$\Rightarrow (x+8)(x-3) < 0$$
$$\Rightarrow -8

Now combining the two results i.e. $$x \ge -33$$ and $$-8 , the answer should be $$-8 as per my understanding but this is not the correct answer that has been provided. What am I doing wrong? Please help me on this !!!

• The issue is here: $x+3<\sqrt{x+33} \Rightarrow(x+3)^2<{x+33}$
– user
Aug 29, 2021 at 16:52

HINT

You can only square both sides iff both sides and the argument of the square root are nonnegative.

Under such conditions, the proposed inequation is equivalent to \begin{align*} x + 3 < \sqrt{x + 33} & \Longleftrightarrow \begin{cases} (x+3)^{2} < x + 33\\\\ x+3 \geq 0\\\\ x + 33 \geq 0 \end{cases}\\\\ & \Longleftrightarrow \begin{cases} x^{2} + 6x + 9 < x + 33\\\\ x\geq -3 \end{cases}\\\\ & \Longleftrightarrow \begin{cases} x^{2} + 6x - 34 < 0\\\\ x \geq -3 \end{cases} \end{align*}

Moreover, notice whenever $$-33 \leq x \leq -3$$, the LHS is negative and the RHS is positive.

Hence the interval $$[-33,-3]$$ is also a solution.

Can you take it from here?

• The correct answer is $-33<x<3$. Aug 29, 2021 at 17:10
• Are you sure? Because $-33 + 3 = -30 < 0 = \sqrt{-33 + 33}$. Similarly, $-3 + 3 = 0 < \sqrt{-3 + 33} = \sqrt{30}$. Aug 29, 2021 at 17:13
• wolframalpha.com/input/?i=x%2B3%3Csqrt%28x%2B33%29 Aug 29, 2021 at 17:13

For the inequality $$\sqrt A> B$$ we need of course $$A\ge 0$$, and you did it correct, but before squaring both sides we need to consider two cases:

• when $$B<0$$ the inequality is always satisfied
• when $$B\ge 0$$ we can square both sides to find other solutions

and you went wrong in that second step.

Therefore in general we have:

$$\sqrt A> B \iff \begin{cases}A\ge 0\\B<0 \end{cases}\quad \lor \quad \begin{cases}A\ge 0\\B>0\\A>B^2 \end{cases}$$

• So I'll have to take these two cases and come up with the solutions for both the cases and then combine the solution, right? Aug 29, 2021 at 17:11
• @Ganit Yes the solution is the union of the solutions for the two systems.
– user
Aug 29, 2021 at 17:13
• thanks...i got the correct answer. thanks for all the help. Aug 29, 2021 at 17:28
• @Ganit Well done! The most important thing is that you undestand the reasoning process under that and why we need to consider the two cases. You are welcome! Bye
– user
Aug 29, 2021 at 17:29

The problem is when you square. Notice that $$x+3 \leq 0$$ whenever $$x \leq -3$$. Also, $$\sqrt{x+33} \geq 0$$ for all values of $$x \geq -33$$. You should take this into consideration when evaluating the inequality.