The problem has two parts: (14) Find the square root of the irrational number $8 + 2\sqrt{15}$. (15) Perform the indicated operations with complex numbers: (a) $(3+5i) + (7-i) + (6+2i)$ (b) $(3+5i) - (7-i) - (6+2i)$ (c) $(3+i\sqrt{2}) + (-2+i\sqrt{8})$

1. Problem Description

The problem has two parts:

(14) Find the square root of the irrational number $8 + 2\sqrt{15}$.

(15) Perform the indicated operations with complex numbers:

(a) $(3+5i) + (7-i) + (6+2i)$

(b) $(3+5i) - (7-i) - (6+2i)$

(c) $(3+i\sqrt{2}) + (-2+i\sqrt{8})$

2. Solution Steps

(14) We are looking for a number $x$ such that $x^2 = 8 + 2\sqrt{15}$.

Assume that $x = a + b$, where $a$ and $b$ are real numbers and we want to find $a$ and $b$ such that $(a+b)^2 = 8 + 2\sqrt{15}$.

$(a+b)^2 = a^2 + 2ab + b^2 = a^2 + b^2 + 2ab$

We want $a^2 + b^2 = 8$ and $2ab = 2\sqrt{15}$, so $ab = \sqrt{15}$.

$a^2b^2 = 15$, and we know $a^2+b^2 = 8$.

Let $u=a^2$ and $v=b^2$. Then $u+v=8$ and $uv=15$.

$u(8-u)=15$, so $8u-u^2=15$, or $u^2-8u+15=0$.

$(u-3)(u-5)=0$, so $u=3$ or $u=5$.

If $u=3$, then $v=5$. If $u=5$, then $v=3$.

Without loss of generality, let $a^2=5$ and $b^2=3$. Then $a=\sqrt{5}$ and $b=\sqrt{3}$.

So, $x = \sqrt{5} + \sqrt{3}$.

$(\sqrt{5} + \sqrt{3})^2 = 5 + 2\sqrt{15} + 3 = 8 + 2\sqrt{15}$.

(15a) $(3+5i) + (7-i) + (6+2i) = (3+7+6) + (5-1+2)i = 16 + 6i$

(15b) $(3+5i) - (7-i) - (6+2i) = 3+5i -7+i -6-2i = (3-7-6) + (5+1-2)i = -10 + 4i$

(15c) $(3+i\sqrt{2}) + (-2+i\sqrt{8}) = 3+i\sqrt{2} -2+i\sqrt{4*2} = 3+i\sqrt{2} -2+2i\sqrt{2} = (3-2) + (1+2)i\sqrt{2} = 1 + 3i\sqrt{2}$

3. Final Answer

(14) $\sqrt{5} + \sqrt{3}$

(15a) $16 + 6i$

(15b) $-10 + 4i$

(15c) $1 + 3i\sqrt{2}$