The problem consists of several math questions. I will answer the following questions: 1. i) If $\sqrt{72} + \sqrt{32} - 3\sqrt{18} = x\sqrt{2}$, find the value of $x$.

1. Problem Description

The problem consists of several math questions. I will answer the following questions:

1. i) If $\sqrt{72} + \sqrt{32} - 3\sqrt{18} = x\sqrt{2}$, find the value of $x$.

2. ii) Convert $413_7$ to a numeral in base

5.

3. i) Simplify $\frac{15}{\sqrt{75}} + (\sqrt{108} + \sqrt{432})$, leaving the answer in the form $a\sqrt{b}$, where $a$ and $b$ are positive integers.

4. ii) Find the quadratic equation whose roots are $-2q$ and $5q$.

5. i) Solve the equation $4x^2 - 16x + 15 = 0$.

6. ii) If $2N4_{seven} = 15N_{nine}$, find the value of $N$.

7. i) The truth set of $8 + 2x - x^2 = 0$ is $\{p, q\}$. Evaluate $p + q$.

8. ii) Given that $\frac{\sqrt{3} + \sqrt{5}}{\sqrt{5}} = x + y\sqrt{15}$, find the value of $(x + y)$.

2. Solution Steps

1. i)

$\sqrt{72} + \sqrt{32} - 3\sqrt{18} = x\sqrt{2}$

$\sqrt{36 \cdot 2} + \sqrt{16 \cdot 2} - 3\sqrt{9 \cdot 2} = x\sqrt{2}$

$6\sqrt{2} + 4\sqrt{2} - 3(3\sqrt{2}) = x\sqrt{2}$

$6\sqrt{2} + 4\sqrt{2} - 9\sqrt{2} = x\sqrt{2}$

$(6 + 4 - 9)\sqrt{2} = x\sqrt{2}$

$1\sqrt{2} = x\sqrt{2}$

$x = 1$

2. ii)

Convert $413_7$ to base 10:

$413_7 = 4 \cdot 7^2 + 1 \cdot 7^1 + 3 \cdot 7^0 = 4 \cdot 49 + 1 \cdot 7 + 3 \cdot 1 = 196 + 7 + 3 = 206_{10}$

Now, convert $206_{10}$ to base 5:

$206 \div 5 = 41 R 1$

$41 \div 5 = 8 R 1$

$8 \div 5 = 1 R 3$

$1 \div 5 = 0 R 1$

Reading the remainders upwards, we get $1311_5$.

3. i)

$\frac{15}{\sqrt{75}} + (\sqrt{108} + \sqrt{432})$

$\frac{15}{\sqrt{25 \cdot 3}} + (\sqrt{36 \cdot 3} + \sqrt{144 \cdot 3})$

$\frac{15}{5\sqrt{3}} + (6\sqrt{3} + 12\sqrt{3})$

$\frac{3}{\sqrt{3}} + 18\sqrt{3}$

$\frac{3\sqrt{3}}{3} + 18\sqrt{3}$

$\sqrt{3} + 18\sqrt{3}$

$19\sqrt{3}$

4. ii)

Let the roots be $-2q$ and $5q$.

Sum of roots: $-2q + 5q = 3q$

Product of roots: $-2q \cdot 5q = -10q^2$

The quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$

$x^2 - (3q)x - 10q^2 = 0$

5. i)

$4x^2 - 16x + 15 = 0$

$4x^2 - 10x - 6x + 15 = 0$

$2x(2x - 5) - 3(2x - 5) = 0$

$(2x - 3)(2x - 5) = 0$

$2x - 3 = 0$ or $2x - 5 = 0$

$2x = 3$ or $2x = 5$

$x = \frac{3}{2}$ or $x = \frac{5}{2}$

6. ii)

$2N4_{seven} = 15N_{nine}$

$2 \cdot 7^2 + N \cdot 7^1 + 4 \cdot 7^0 = 1 \cdot 9^2 + 5 \cdot 9^1 + N \cdot 9^0$

$2 \cdot 49 + 7N + 4 = 1 \cdot 81 + 5 \cdot 9 + N$

$98 + 7N + 4 = 81 + 45 + N$

$102 + 7N = 126 + N$

$6N = 24$

$N = 4$

7. i)

$8 + 2x - x^2 = 0$

$x^2 - 2x - 8 = 0$

$(x - 4)(x + 2) = 0$

$x = 4$ or $x = -2$

Truth set $\{p, q\} = \{4, -2\}$

$p + q = 4 + (-2) = 2$

8. ii)

$\frac{\sqrt{3} + \sqrt{5}}{\sqrt{5}} = x + y\sqrt{15}$

$\frac{\sqrt{3}}{\sqrt{5}} + \frac{\sqrt{5}}{\sqrt{5}} = x + y\sqrt{15}$

$\frac{\sqrt{3}}{\sqrt{5}} + 1 = x + y\sqrt{15}$

$\frac{\sqrt{3}\sqrt{5}}{\sqrt{5}\sqrt{5}} + 1 = x + y\sqrt{15}$

$\frac{\sqrt{15}}{5} + 1 = x + y\sqrt{15}$

$1 + \frac{1}{5}\sqrt{15} = x + y\sqrt{15}$

$x = 1$ and $y = \frac{1}{5}$

$x + y = 1 + \frac{1}{5} = \frac{6}{5}$

3. Final Answer

1. i) $x = 1$

2. ii) $413_7 = 1311_5$

3. i) $19\sqrt{3}$

4. ii) $x^2 - 3qx - 10q^2 = 0$

5. i) $x = \frac{3}{2}, \frac{5}{2}$

6. ii) $N = 4$

7. i) $p + q = 2$

8. ii) $x + y = \frac{6}{5}$