• ## Is this a thing?

1 reply omitted in this preview.

• Paywalled.
Can someone with access to the article, summarize shat the trick is ?

26°
• >>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."

28°
• >>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.

29°
• 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.

• >>3
maybe not? but, you know that's a really hard problem, right?....
do you even know anything about math?

13°
• The g-ds never give mortals possible tasks. Otherwise, they would have done it themselves.

19°
• 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.

27°
• ## 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.

15°
• ## For all e>0, there exists a d>0 such that...

• Did you know that you can use LaTeX on this website?

$forall \ \varepsilon > 0, \exists \delta > 0$ such that...
10°
• >>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°
• >>11
Of course, it doesn't work for post titles.

12°
• ## Hello!

I love math!

• I love math too!