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## 离散数学代写

A quadratic equation is an equation of the form $a x^{2}+b x+c=0$, and solving the quadratic equation is concerned with finding the unknown value $x$ (roots of the quadratic equation). There are several techniques to solve quadratic equations such as factorization; completing the square, the quadratic formula and graphical techniques.
Solve the quadratic equation $3 x^{2}-11 x-4=0$ by factorization.
The approach taken is to find the factors of the quadratic equation. Sometimes this is easy, but often other techniques will need to be employed. For the above quadratic equation, we note immediately that its factors are $(3 x+1)(x-4)$ since
\begin{aligned} &(3 x+1)(x-4) \ &=3 x^{2}-12 x+x-4 \ &=3 x^{2}-11 x-4 \end{aligned}
Next, we note the property that if the product of two numbers $A$ and $B$ is 0 then either $A$ is 0 or $B$ is 0 . In other words, $A B=0=>A=0$ or $B=0$. We conclude from this property that as
$$3 x^{2}-11 x-4=0$$
• $(3 x+1)(x-4)=0$

• $(3 x+1)=0$ or $(x-4)=0$
• $3 x=-1$ or $x=4$
• $x=-0.33$ or $x=4$.
Therefore, the solution (or roots) of the quadratic equation $3 x^{2}-11 x-4=0$ is $x=-0.33$ or $x=4$ \begin{tabular}{l} Example (Quadratic Equations-Completing the Square) \ Solve the quadratic equation $2 x^{2}+5 x-3=0$ by completing the square. \ Solution (Quadratic Equations-Completing the Square) \ First, we convert the quadratic equation to an equivalent quadratic with a unary \ coefficient of $x 2$. This involves division by 2 . Next, we examine the coefficient of $x$ \ (in this case $5 / 2)$, and we add the square of half the coefficient of $x$ to both sides. \ This allows us to complete the square, and we then take the square root of both \ sides. Finally, we solve for $x$. \ 108 \ \hline \end{tabular}
• $x^{2}+5 / 2 x-3 / 2=0$
• $x^{2}+5 / 2 x=3 / 2$
• $x^{2}+5 / 2 x+(5 / 4)^{2}=^{3} / 2+(5 / 4)^{2}$
• $(x+5 / 4)^{2}=^{3} / 2+\left({ }^{25} / 16\right)$
• $(x+5 / 4)^{2}=^{24} / 16+(25 / 16)$
• $(x+5 / 4)^{2}={ }^{49} / 16$
• $(x+5 / 4)=\pm^{7} / 4$
• $x=-5 / 4^{5} / 4$
• $x=-5 / 4-{ }^{7} / 4$ or $x=-5 / 4+7 / 4$
• $x=-{ }^{12} /{ }_{4}$ or $x={ }^{2} / 4$
• $x=-3$ or $x=0.5$.

$$\开始{对齐} &(3 x+1)(x-4) \ &=3 x^{2}-12 x+x-4 \ &=3 x^{2}-11 x-4 \end{对齐}$$

$$3 x^{2}-11 x-4=0$$
• $(3 x+1)(x-4)=0$

• $(3 x+1)=0$ 或 $(x-4)=0$
• $3 x=-1$ 或 $x=4$
• $x=-0.33$ 或 $x=4$。
因此，二次方程 $3 x^{2}-11 x-4=0$ 的解（或根）为 $x=-0.33$ 或 $x=4$ \begin{tabular}{l} 示例（二次Equations-Completing the Square) \ 通过完成平方求解二次方程 $2 x^{2}+5 x-3=0$。 \解（二次方程-完成平方）\首先，我们将二次方程转换为具有一元\系数$x 2$的等价二次方程。这涉及除以 2 。接下来，我们检查$x$ \ (在本例中为$5 / 2)$ 的系数，并将$x$ 系数一半的平方加到两边。 \ 这使我们可以完成平方，然后我们取 \ 两边的平方根。最后，我们求解 $x$。 \ 108 \ \hline \end{表格}
• $x^{2}+5 / 2 x-3 / 2=0$
• $x^{2}+5 / 2 x=3 / 2$
• $x^{2}+5 / 2 x+(5 / 4)^{2}=^{3} / 2+(5 / 4)^{2}$
• $(x+5 / 4)^{2}=^{3} / 2+\left({ }^{25} / 16\right)$
• $(x+5 / 4)^{2}=^{24} / 16+(25 / 16)$
• $(x+5 / 4)^{2}={ }^{49} / 16$
• $(x+5 / 4)=\pm^{7} / 4$
• $x=-5 / 4^{5} / 4$
• $x=-5 / 4-{ }^{7} / 4$ 或 $x=-5 / 4+7 / 4$
• $x=-{ }^{12} /{ }_{4}$ 或 $x={ }^{2} / 4$
• $x=-3$ 或 $x=0.5$。

## 图论代考

(a) 假设一个操作有 $m$ 个可能的结果，而第二个操作有 $n$ 个可能的结果，那么执行第一个操作后执行第二个操作时可能结果的总数是 $m \times n$ (Product Rule )。
(b) 假设一个操作有 $m$ 个可能的结果，而第二个操作有 $n$ 个可能的结果，那么第一个操作或第二个操作的可能结果总数由 $m+n$ 给出（求和规则） .

$(1, \mathrm{H}),(2, \mathrm{H}),(3, \mathrm{H}),(4, \mathrm{H}),(5, \mathrm{H}) ,(6, \mathrm{H}),(1, \mathrm{~T}),(2, \mathrm{~T}),(3, \mathrm{~T}),(4, \mathrm{ ~T}),(5, \mathrm{~T}),(6, \mathrm{~T})$

5.7 排列组合
97

(a) 假设有一组 367 人，那么必须至少有两个人的生日相同。

(b) 假设有 102 名学生参加了一次考试（考试的结果是 0 到 100 之间的分数）。然后，至少有两名学生获得相同的分数。

## 密码学代考

• Cryptosystem
• A system that describes how to encrypt or decrypt messages
• Plaintext
• Message in its original form
• Ciphertext
• Message in its encrypted form
• Cryptographer
• Invents encryption algorithms
• Cryptanalyst
• Breaks encryption algorithms or implementations

## 编码理论代写

1. 数据压缩（或信源编码
2. 前向错误更正（或信道编码
3. 加密编码
4. 线路码