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Actually this can't be solved by making it a cube.it can't be made because then b and c cannot be arbitrary.there must exist a relation like $ b^2/3.a^2 = c/a $. ; if you differentiate, you will get that the cubic equation having only 1 root has exactly that condition.

I would very much like to have a complete list of the types of polynomial functions. I know that: \\begin{align*} \\text{Quadratic}: \\qquad & ax^2+bx+c ...

2018年10月11日 · If you were to graph points and try to solve the general cubic graphically, you would iteratively calculate points until you found the place that the curve crossed zero -- guess and check for every cubic you needed to solve.

$\begingroup$ There are only two square roots of ii (as there are two square roots of any non-zero complex number), namely $\pm(1+i)/\sqrt{2}$.

2017年12月25日 · I may first give an example : finding limit $$ \lim_{x \rightarrow \infty} \frac{1+x}{x} $$ When we use straightforward approach, we get $$ \frac{\infty+1}{\infty} = \frac{\infty}{\infty} $$ In the process of investigating a limit, we know that both the numerator and denominator are going to infinity.. but we dont know the behaviour of each dynamics.

I am aware that, historically, hyperbolic geometry was useful in showing that there can be consistent geometries that satisfy the first 4 axioms of Euclid's elements but not the fifth, the infamous

2014年10月6日 · How do you prove the following: Pythagorean trigonometric identity. For all $\theta\in[0,2\pi]$ it holds that $$ \sin^2\theta+\cos^2\theta=1.$$ I'm curious to know of the different ways of provin...

2017年6月19日 · A bit more general : 1) Consider a family of planes with normalized (length = 1) normal vectors $\vec{n} $.

$\begingroup$ Forgive my ignorance, but isn't it a little simpler than this? Here's my argument. Since $2^2 < (\sqrt{5})^2 < 3^2,$ and since the positive square root function is strictly increasing, thus $2 < \sqrt{5} < 3.$ Since there are not natural numbers between $2$ and $3$, this means that $\sqrt{5}$ is non-natural.