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

Question

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

Now combining the two results i.e. $ x \ge -33$ and $ -8<x<3$ , the answer should be $ -8<x<3$ 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 !!!

Thanks in advance !!!

Answer 1

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?

Answer 2

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:

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

Answer 3

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.