We need to solve two inequalities involving square roots. (g) $\sqrt{2x+5} < \sqrt{9+x}$ (h) $\sqrt{x+3} + \sqrt{x+7} > 4$

1. Problem Description

We need to solve two inequalities involving square roots.

(g) $\sqrt{2x+5} < \sqrt{9+x}$

(h) $\sqrt{x+3} + \sqrt{x+7} > 4$

2. Solution Steps

(g) $\sqrt{2x+5} < \sqrt{9+x}$

First, we need to ensure that the expressions under the square roots are non-negative. Thus, we have:

$2x+5 \ge 0 \implies x \ge -\frac{5}{2}$

$9+x \ge 0 \implies x \ge -9$

So, we have $x \ge -\frac{5}{2}$.

Now, square both sides of the inequality:

$(\sqrt{2x+5})^2 < (\sqrt{9+x})^2$

$2x+5 < 9+x$

$2x-x < 9-5$

$x < 4$

Combining the two conditions, we have $-\frac{5}{2} \le x < 4$.

(h) $\sqrt{x+3} + \sqrt{x+7} > 4$

First, we need to ensure that the expressions under the square roots are non-negative. Thus, we have:

$x+3 \ge 0 \implies x \ge -3$

$x+7 \ge 0 \implies x \ge -7$

So, we have $x \ge -3$.

Now, isolate one of the square roots:

$\sqrt{x+7} > 4 - \sqrt{x+3}$

We consider two cases:

Case 1: $4 - \sqrt{x+3} < 0$, which means $\sqrt{x+3} > 4$.

$x+3 > 16$

$x > 13$

Since $\sqrt{x+7}$ is always positive, $\sqrt{x+7} > 4 - \sqrt{x+3}$ always holds for $x > 13$.

Case 2: $4 - \sqrt{x+3} \ge 0$, which means $\sqrt{x+3} \le 4$, or $x+3 \le 16$, so $x \le 13$. Since $x \ge -3$, we have $-3 \le x \le 13$.

Square both sides of the inequality:

$(\sqrt{x+7})^2 > (4 - \sqrt{x+3})^2$

$x+7 > 16 - 8\sqrt{x+3} + x+3$

$x+7 > x+19 - 8\sqrt{x+3}$

$8\sqrt{x+3} > 12$

$\sqrt{x+3} > \frac{12}{8} = \frac{3}{2}$

$x+3 > \frac{9}{4}$

$x > \frac{9}{4} - 3 = \frac{9-12}{4} = -\frac{3}{4}$

Combining with $-3 \le x \le 13$, we have $-\frac{3}{4} < x \le 13$.

Combining Case 1 and Case 2, we have $x > -\frac{3}{4}$.

3. Final Answer

(g) $-\frac{5}{2} \le x < 4$

(h) $x > -\frac{3}{4}$