338 votes
Accepted

If squaring a number means multiplying that number with itself then shouldn't taking square root of a number mean to divide a number by itself?

taking square root means reversing the effect of squaring. Dividing a number by itself does not do that (but rather always ...
dxiv's user avatar
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272 votes
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If there are $74$ heads and $196$ legs, how many horses and humans are there?

A hypercentaur is a creature with $2$ heads and $6$ legs; an anticentaur is a creature with no head and $-2$ legs. Since $74$ heads make for $37$ hypercentaurs, with $74\cdot 6/2=222$ legs, you have $(...
egreg's user avatar
  • 237k
190 votes
Accepted

Why can a quadratic equation have only 2 roots?

Suppose there are three distinct roots $x,y,z$. One has $$\begin{cases}ax^2+bx+c=0\\ay^2+by+c=0\\az^2+bz+c=0\end{cases}\Rightarrow\begin{cases}a(x^2-y^2)+b(x-y)=0\\a(x^2-z^2)+b(x-z)=0\end{cases}\...
Piquito's user avatar
  • 28.7k
168 votes
Accepted

Why do I get one extra wrong solution when solving $2-x=-\sqrt{x}$?

This is because the equation $\;\sqrt x=x-2$ is not equivalent to $x=(x-2)^2$, but to $$x=(x-2)^2\quad\textbf{and}\quad x\ge 2.$$ Remember $\sqrt x$, when it is defined, denotes the non-negative ...
Bernard's user avatar
  • 175k
167 votes

Why is negative times negative = positive?

This is pretty soft, but I saw an analogy online to explain this once. If you film a man running forwards ($+$) and then play the film forward ($+$) he is still running forward ($+$). If you play ...
miradulo's user avatar
  • 3,762
165 votes
Accepted

Why does $\frac{1}{x} < 4$ have two answers?

You have to be careful when multiplying by $x$ since $x$ might be negative and hence flip the inequality. Suppose $x>0$. Then $$\frac{1}{x}<4\iff4x>1\iff x>1/4.$$ If $x>0$ and $x>1/4$...
Sri-Amirthan Theivendran's user avatar
151 votes
Accepted

Multiple-choice: sum of primes below $1000$

The sum of the first 168 positive integers is $\frac{168^2+168}{2}=14196$, which is greater than answer (a). The sum of the first 168 primes must be even greater than that.
Meni Rosenfeld's user avatar
141 votes

In a village, $90\%$ of people drink Tea, $80\%$ Coffee, $70\%$ Whiskey, $60\%$ Gin. Nobody drinks all four. What percentage of people drinks alcohol?

If you add up the percentages, they come out to $300\%$. This means that the average number of beverages per person is $3$. No one drinks more than that, so no one can drink less than that, either. ...
Acccumulation's user avatar
133 votes

Why does the discriminant in the Quadratic Formula reveal the number of real solutions?

Think about it geometrically $-$ then compute. Everyone knows $x^2$ describes a parabola with its apex at $(0,0)$. By adding a parameter $\alpha$, we can move the parabola up and down: $x^2+\alpha$ ...
M. Winter's user avatar
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122 votes

Why does $\frac{1}{x} < 4$ have two answers?

Here is the solution $$\frac { 1 }{ x } <4$$$$ \frac { 1-4x }{ x } <0$$$$ \frac { x\left( 1-4x \right) }{ { x }^{ 2 } } <0$$$$ x\left( 1-4x \right) <0$$$$ x\left( 4x-1 \right) >0 $$ ...
haqnatural's user avatar
  • 21.5k
116 votes

What actually is a polynomial?

A polynomial (in one variable) is an expression of the form $$ p(x) = a_0+a_1x+a_2x^2+\ldots+a_nx^n$$ where the coefficients $a_i$ are some kind of number (or more generally they're elements of a Ring)...
spaceisdarkgreen's user avatar
115 votes

Find the height of a bar, given the lengths of shadows cast by it and another bar

Here is a different visualization.
Doug M's user avatar
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103 votes
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Solution to the equation of a polynomial raised to the power of a polynomial.

Denote $a=x^2-7x+11.$ The equation becomes $a^{a-5}=1,$ or equivalently* $$a^a=a^5,$$ which has in $\mathbb{R}$ the solutions $a\in \{ {5,1,-1}\}.$ Solving the corresponding quadratic equations we get ...
user376343's user avatar
  • 8,236
100 votes

What is the smallest integer greater than 1 such that $\frac12$ of it is a perfect square and $\frac15$ of it is a perfect fifth power?

The number is clearly a multiple of $5$ and $2$. We look for the smallest, so we assume that it has no more prime factors. So let $n=2^a5^b$. Since $n/2$ is a square, then $a-1$ and $b$ are even. ...
ajotatxe's user avatar
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96 votes
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Can I think of Algebra like this?

My question is: Is there any disadvantages to thinking about Algebra like this? Is there anything later in my math education that will require me to know that I am subtracting or adding 2x to get ...
Daniel R. Collins's user avatar
94 votes

Why does cancelling change this equation?

When you cancel out $t$ on both sides you are assuming $t\neq0$. To be rigorous you need to divide into two cases, $t\neq0$, then you can cancel out $t$. $t=0$, then you cannot cancel out $t$ as ...
cr001's user avatar
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93 votes
Accepted

Systems of linear equations: Why does no one plug back in?

You wrote this step as an implication: $$\begin{cases} -2x-y=0 \\ 3x+y=4 \end{cases} \implies \begin{cases} -2x-y=0\\ x=4 \end{cases}$$ But it is in fact an equivalence: $$\begin{cases} -2x-y=0 \...
dxiv's user avatar
  • 76.2k
93 votes

$r=\pm1$ are the only rationals with $\,r+1/r\in \Bbb Z$ (sum with its reciprocal is an integer)

It seems like you are asking for a rational number $n$ with the property that $$n+\frac{1}{n}$$ is an integer. Let $z$ be an integer. Then we have $$n+\frac{1}{n}=z$$ and $$n^2+1=zn$$ $$n^2-zn+1=0$$ ...
Franklin Pezzuti Dyer's user avatar
91 votes

Is there a clever solution to Arnold's "merchant problem"?

At the end the tea cup is as full as at the start. This implies that the added wine is exactly outweighed by the tea that has disappeared.
Christian Blatter's user avatar
90 votes
Accepted

What is the definition of a set?

Formally speaking, sets are atomic in mathematics.1 They have no definition. They are just "basic objects". You can try and define a set as an object in the universe of a theory designated ...
Asaf Karagila's user avatar
  • 390k
88 votes
Accepted

Are polynomials with the same roots identical?

No, they are not. For instance, $2x^2-2$ and $x^2-1$ have the same roots, yet they are not identical. And, depending on what you mean by "the same roots", we have that $x^2-2x+1$ and $x-1$ have the ...
Arthur's user avatar
  • 197k
87 votes

Do there exist pairs of distinct real numbers whose arithmetic, geometric and harmonic means are all integers?

Expanding on Christian Blatter's answer. There are a few key points. The arithmetic mean of two rational numbers is always rational. The harmonic mean of two non-zero rational numbers is always ...
Peter Green's user avatar
  • 1,256
86 votes
Accepted

Is $\sqrt{64}$ considered $8$? or is it $8,-8$?

Your new teacher is wrong. $\sqrt{\cdot}$ is the principal square root operator. That means it returns only the principal root -- the positive one. $\sqrt{64}=8$. It does NOT equal $-8$. On the ...
got it--thanks's user avatar
85 votes
Accepted

Mathematical symbol for 'slightly greater than'?

More often it is used as $b=a+\epsilon$ where $\epsilon$ normally stands for a small positive quantity. That provides b slightly greater than a. Similarly $-\epsilon$ for slightly below.
AHusain's user avatar
  • 5,044
83 votes

Intuition for why the difference between $\frac{2x^2-x}{x^2-x+1}$ and $\frac{x-2}{x^2-x+1}$ is a constant?

Would you be surprised that the difference of $\dfrac{2x^2+x+1}{x^2}$ and $\dfrac{x+1}{x^2}$ is $2$?
Jared Goguen's user avatar
  • 1,087
81 votes

Why $\sqrt{-1 \cdot {-1}} \neq \sqrt{-1}^2$?

The simple reason this, $$\sqrt{\frac{-1}{1}} =\frac{\sqrt{1}}{\sqrt{-1}}$$ is not valid is because of a branch cut that must be taken as a result of the square root function. Using exponentials: $$ \...
Alexander McFarlane's user avatar
81 votes
Accepted

What is an intuitive approach to solving $\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$?

Intuition should say: the denominator grows with $n^2$, the numerator grows with $n$. However, the number of fractions also grows by $n$, so the total growth of the numerator is about $n^2$. And ...
5xum's user avatar
  • 122k
80 votes

What actually is a polynomial?

There are lots of good answers here and they are all essentially correct, even though they are different! I will try to contribute another, which is somewhat more abstract than the others. I normally ...
Ethan Bolker's user avatar
  • 91.8k

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