Is this a thing?
https://www.nytimes.com/2020/02/05/science/quadratic-equations-algebra.html
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Paywalled.
Can someone with access to the article, summarize shat the trick is ? -
>>26
"This alternate method for solving quadratic equations uses the fact that parabolas are symmetrical.For example, in this parabola:
y = x^2 – 4x – 5
The two solutions when y = 0 are the symmetrical points r and s, where the parabola crosses the x-axis.
The midpoint, or average, of r and s is the axis of symmetry of the parabola. We want r + s = –b, which happens when the average of r and s is –b ÷ 2. In this example: 4 ÷ 2 = 2.
The two solutions to the quadratic equation will be the axis of symmetry plus or minus an unknown amount, which we’ll call u. In this example:
r = 2 – u and s = 2 + u
To find u, we want the product of r and s to be equal to c, which in this example is –5. Rewriting r and s in terms of u:
r × s = –5
(2 – u) × (2 + u) = –5
Solving that yields 22 – u2 = –5 or u2 = 9, so u = 3 works.
The two solutions to this quadratic equation are 2 – u and 2 + u, or –1 and 5. In other words, this parabola intersects the x-axis when x = –1 and x = 5."
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>>28
>Solving that yields 22 – u2 = –5 or u2 = 9, so u = 3 works.
Solving that yields 22 – u^2 = –5 or u^2 = 9, so u = 3 works.Sorry. Copy-and-paste mistake.
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Ever since that other anonymous website made a major contribution to mathematics, I feel like we should be able to do it too. I vote that we work on the Riemann Hypothesis.
2 replies omitted in this preview.
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>>3
maybe not? but, you know that's a really hard problem, right?....
do you even know anything about math? -
The g-ds never give mortals possible tasks. Otherwise, they would have done it themselves.
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Shouldn't we opt for something simpler first? Like developing calculus without limits or infinities - and then use that for something trivial, like plotting trajectory of a planet moving under influence of a black hole? Of course, somewhere in between we'll need to develop new theory of gravity with our new, finite, calculus.
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The square root of 2 is irrational
Let $x$ be a positive number such that $x^2 = 2$. Assume for the sake of contradiction that $x$ is rational. Then there exist coprime non-negative integers $a,b$, with $b \ne 0$, such that $x = \frac{a}{b}$. Then we have $2 = x^2 = \frac{a^2}{b^2}$, so $a^2 = 2b^2$. Since $2$ is prime and $2$ divides $a^2$, $2$ must divide $a$. Say $a = 2c$ for some integer $c$. Then $4c^2 = a^2 = 2b^2$, so $2c^2 = b^2$, and therefore $2$ divides $b$. But then $a$ and $b$ are not coprime, which is a contradiction. Therefore $x$ is irrational. QED-
Thank you for your wonderful contribution! Always love to see some real maths here.
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For all e>0, there exists a d>0 such that...
e-d statements are rad
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Did you know that you can use LaTeX on this website?
$forall \ \varepsilon > 0, \exists \delta > 0$ such that... -
>>10
Although, not all things you'd expect are supported so I guess my forall symbol failed there.Anyway, just type it like this.
%%% some text $some latex$ more text %%% -
>>11
Of course, it doesn't work for post titles.
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Hello!
I love math!
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I love math too!
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What's your favourite theorem?
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Well, it would have to be Fermat's Last, wouldn't it.......
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Ummm, is it normal to know a lot of these theorems offhand?
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