2017年9月19日 · $$(\alpha-\beta)^2 = (\alpha+\beta)^2-4\alpha \beta = \dfrac{4pr +q^2}{p^2} $$ $$ \alpha -\beta =\pm \dfrac{\sqrt{ 4pr +q^2}}{{p}}$$ Actually you can write out the qudratic roots separately and subtract one from the other.. even if it appears brute force. The discriminant is an important part of the result. (It vanishes for equal roots).

2013年8月19日 · My working out so far: I know that $\alpha + \beta = -8$ and $\alpha \beta = -5$ (from the roots) and thenIi go on to work out that $\alpha= -8-\beta$ and $\beta= -8-\alpha$, then I substitute into what the question asks me. $\frac{-8-\beta}{-8-\alpha}$ and $\frac{-8-\alpha}{-8-\beta}$ however I do not know how to proceed further.

2016年8月30日 · In order to do that I can set "alfa" and "beta" angles in the motor so they move the arm to the position I want. The problem is that I came up with 2 equations and 2 variables (alfa ($\alpha$) and beta($\beta$)) and I tried really hard but I still can't isolate $\alpha$ and $\beta$ so I can calculate it's value directly, using a microcontoller.

Now if you believe that rotations are linear maps and that a rotation by an angle of $\alpha$ followed by a rotation by an angle of $\beta$ is the same as a rotation by an angle of $\alpha+\beta$ then you are lead to \begin{align} D_{\alpha+\beta}&=D_\beta D_\alpha, & D_\phi&=\begin{pmatrix} \cos\phi&-\sin\phi\\ \sin\phi&\cos\phi \end{pmatrix ...

2024年10月28日 · Quadratic Equation whose roots are $\alpha$ and $\beta$ It is given that roots of $(a+b+c)x^2+(b+2c)x+c=0$ are $\frac{\alpha}{\alpha -1}$ and $\frac{\beta}{\beta -1}$. Now we have to find the quadratic equation whose roots are $\alpha$ and $\beta$ without finding $\alpha + \beta$ and $\alpha \beta$. I took

2019年9月18日 · If $$ \\alpha + \\beta + \\gamma = 14 $$ $$ \\alpha^{2} + \\beta^{2} + \\gamma^{2} = 84 $$ $$ \\alpha^{3} + \\beta^{3} + \\gamma^{3} = 584 $$ What is $\\alpha^{4 ...

2015年5月14日 · $\cos(n\alpha)$ is indeed expressible as a polynomial in terms of $\cos(\alpha)$ (and $\sin(\alpha)$). Reciprocally, you can in some cases solve that polynomial to obtain $\cos(\alpha/n)$ in terms of $

2019年3月31日 · I drew a simple diagram from which one can deduce that $$\alpha=10^\circ$$ $$\beta=40^\circ-\alpha=30 ...

2016年4月25日 · Yes, it is most definitely valid for $\alpha < \beta$. This is simple to show. Assume it's correct for $\alpha > \beta$:

2021年7月10日 · While reading about quadratic equations, I came across Newton's Identity formula which said we can express $\alpha^n+\beta^n$ in simpler forms but not given any explanation.