We need to solve two equations for real values of $x$. i. $\sqrt{x+2} + \sqrt{2x} = \sqrt{18-x}$ ii. $\frac{x-3}{x^2+2} = \frac{1}{x+1}$

1. Problem Description

We need to solve two equations for real values of $x$.

i. $\sqrt{x+2} + \sqrt{2x} = \sqrt{18-x}$

ii. $\frac{x-3}{x^2+2} = \frac{1}{x+1}$

2. Solution Steps

i. $\sqrt{x+2} + \sqrt{2x} = \sqrt{18-x}$

First, we need to consider the domain of the variable $x$.

$x+2 \ge 0 \implies x \ge -2$

$2x \ge 0 \implies x \ge 0$

$18-x \ge 0 \implies x \le 18$

Combining these, we have $0 \le x \le 18$.

Square both sides of the equation:

$(\sqrt{x+2} + \sqrt{2x})^2 = (\sqrt{18-x})^2$

$(x+2) + 2\sqrt{(x+2)(2x)} + 2x = 18-x$

$3x+2 + 2\sqrt{2x^2+4x} = 18-x$

$2\sqrt{2x^2+4x} = 16-4x$

$\sqrt{2x^2+4x} = 8-2x$

Square both sides again:

$2x^2+4x = (8-2x)^2$

$2x^2+4x = 64 - 32x + 4x^2$

$0 = 2x^2 - 36x + 64$

$x^2 - 18x + 32 = 0$

$(x-2)(x-16) = 0$

$x=2$ or $x=16$

Check $x=2$:

$\sqrt{2+2} + \sqrt{2(2)} = \sqrt{18-2}$

$\sqrt{4} + \sqrt{4} = \sqrt{16}$

$2+2 = 4$

$4 = 4$. So $x=2$ is a solution.

Check $x=16$:

$\sqrt{16+2} + \sqrt{2(16)} = \sqrt{18-16}$

$\sqrt{18} + \sqrt{32} = \sqrt{2}$

$3\sqrt{2} + 4\sqrt{2} = \sqrt{2}$

$7\sqrt{2} = \sqrt{2}$, which is false. So $x=16$ is not a solution.

ii. $\frac{x-3}{x^2+2} = \frac{1}{x+1}$

$(x-3)(x+1) = x^2+2$

$x^2 - 3x + x - 3 = x^2 + 2$

$x^2 - 2x - 3 = x^2 + 2$

$-2x - 3 = 2$

$-2x = 5$

$x = -\frac{5}{2}$

3. Final Answer

i. $x = 2$

ii. $x = -\frac{5}{2}$