TITLE: need help solving a quartic equation
QUESTION [0 upvotes]: The question I am asking is to solve the equation $x^4-4x-1=0$,
I need an exact answer. What I have done was found out that it equals $(x^2+1)^2 - 2(x+1)^2 =0$.
Anybody help me, please?

REPLY [1 votes]: HINT
Use $a^2-b^2=(a-b)(a+b)$ to get $2$ quadratic equations$(=0)$.


TITLE: a problem on unimodular matrix
QUESTION [2 upvotes]: A unimodular matrix is a matrix with integer entries and having determinant $1$ or $-1$. If $m$ and $n$ are positive integers, write down a necessary and sufficient condition such that there exists a unimodular matrix of order $2$ whose first row is the vector ($m, n$)

totally stuck.can anybody help.

REPLY [4 votes]: I suspect by "unimodular matrix of order 2" you mean a $2 \times 2$ unimodular matrix.  (Having order 2 means something different than this.)
So let's write down a matrix as specified in the problem:
$$
M = \begin{pmatrix} m & n \\ a & b \end{pmatrix}.
$$
Then $\det(M) = bm - an$.  In order for $bm - an = \pm 1$, $m$ and $n$ cannot have a common factor (why?).
Conversely, suppose that $m$ and $n$ are relatively prime (i.e. have no common factor).  Can we always find integers $a,b$ such that $bm - an = \pm 1$?  Hint: use the Euclidean algorithm.

REPLY [1 votes]: Try $gcd(m,n)=1$. If it is 1 there are ways to solve $mx+ny=1$ in integers $x,y$.
If it is not 1, there is no way to complete the matrix to have unimodular.


TITLE: Writing the density function for a random variable over a selection from a population
QUESTION [0 upvotes]: Problem

A population consists of people who are either blue or not blue. A sample of $5$ people are chosen one at a time, with replacement. Suppose that $\%20$ of the population is blue. Let $X$ be a random variable which gives the number of blue people in the sample. Write the density function, $f_X$, of $X$.

Attempted solution...
Clearly, the support set of $X$ is given by $S = \{0,1,2,3,4,5\}$. For $n\in S$, it seems to me that the probability of $n$ is given by $(0.8^{5 - n})(0.2^n)$, giving the density function:
\begin{cases}
    0.8^5 & x = 0\\
  (0.8^4)(0.2) & x = 1\\
  (0.8^3)(0.2^2) & x = 2\\
  (0.8^2)(0.2^3) & x = 3\\
  (0.8^1)(0.2^4) & x = 4\\
  (0.2^5) & x = 5\\
  \end{cases}
This seems, at first, fairly intuitive. However, summing the probabilities yields $0.4368$, where I'd expect the sum to be $1$. I believe, then, that I've erred somewhere. I'm not sure where, however? Or perhaps, for some reason, I shouldn't expect the sum to be $1$?
What's going on here?

REPLY [0 votes]: $(0.8)^4(0.2)$ is the probability for selecting four non-blue people, followed by one blue person, in that order.
There are $5$ ways to rearrange that ordering.   That is the count for ways to select which position the blue person is placed in a selection.
The probability that there is exactly one blue person (somewhere) among the five selected is $5(0.8)^4(0.2)$

In general, there are $\binom 5k$ ways to select $k$ positions among $5$.
So your table needs to be modified using this Binomial Coefficient.

PS: The Binomial Theorem states $\displaystyle(0.8+0.2)^5=\sum_{k=0}^5 \binom 5k (0.8)^{5-k}(0.2)^k$