Algebra Word Problem Solutions

How to solve a quadratic that needs complex numbers?
${\displaystyle x^2-2x\cos \left(\alpha \right)+1}$

${\displaystyle \frac{1}{\sqrt{7}}÷7=\frac{1}{7\sqrt{7}}=\frac{1}{7\sqrt{7}}\cdot \frac{\sqrt{7}}{\sqrt{7}}=\cdots }$

${\displaystyle (\frac{r}{\sqrt{r^2-1}},\frac{r}{\sqrt{r^2+1}})}$ are roots of the equation ${\displaystyle x^2-bx+3=0}$. What is the value of b?

Proving the limit of ${\displaystyle \frac{{\log \left(n\right)}^{\log \left(n\right)}}{{1.01}^n}}$
Can anyone please show me a simple way (if there is one) to show that
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{{\log \left(n\right)}^{\log \left(n\right)}}{{1.01}^n}=0}$
And that
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{{1.01}^n}{n!}=0}$
I've checked that it's true, I just need to show it the shortest way possible.

Proving roots of quadratic equations
If ${\displaystyle \alpha ,\beta }$ are the roots of the quadratic equation ${\displaystyle a{x}^2+bx+c=0}$, obtain the equation whose roots are ${\displaystyle \frac{1}{{\alpha }^3}\text{and}\frac{1}{{\beta }^3}}$.
If, in the above equation ${\displaystyle \alpha {\beta }^2=1}$, prove that ${\displaystyle a^3+c^3+abc=0}$.

Provide the solution
1)${\displaystyle \log 3(x^2+2)=3}$
2)${\displaystyle {\log }_3(x+4)={\log }_3(2x-4)}$
3)${\displaystyle {\log }_2(3x+5)\ge {\log }_2(x-9)}$
4)${\displaystyle {\log }_4(x+1)<{\log }_4\left(2x\right)}$

Prove this has real roots ${\displaystyle (a^2-bc)x^2+(a+b)(a-c)x+a(b-c)=0}$

Separating ${\displaystyle \frac{1}{1-x^2}}$ into multiple terms
I'm working through an example that contains the following steps:
${\displaystyle \int \frac{1}{1-x^2}dx}$
${\displaystyle =\frac{1}{2}\int \frac{1}{1+x}-\frac{1}{1-x}dx}$
${\displaystyle =\frac{1}{2}\ln \frac{1+x}{1-x}}$
I don't understand why the separation works. If I attempt to re-combine the terms, I get this:
${\displaystyle \frac{1}{1+x}\frac{1}{1-x}}$
${\displaystyle =\frac{1-x}{1-x}\frac{1}{1+x}-\frac{1+x}{1+x}\frac{1}{1-x}}$
${\displaystyle =\frac{1-x-(1+x)}{1-x^2}}$
${\displaystyle =\frac{-2x}{1-x^2}\ne \frac{2}{1-x^2}}$
Or just try an example, and plug in ${\displaystyle x=2}$:
${\displaystyle 2\frac{1}{1-2^2}=\frac{-2}{3}}$
${\displaystyle \frac{1}{1+2}-\frac{1}{1-2}=\frac{1}{3}+1=\frac{4}{3}\ne \frac{-2}{3}}$
Why can ${\displaystyle \frac{1}{1-x^2}}$ be split up in this integral, when the new terms do not equal the old term?

S and G are roots of ${\displaystyle x^2-3x+1=0}$. Find equation whose roots are ${\displaystyle \frac{1}{S-2}}$ and ${\displaystyle \frac{1}{G-2}}$.

Rewrite the rational exponent as a radical by extending the properties of integer exponents.
11 to the two fifths power over 11 to the one fourth power

Region of the coefficients of a quadratic equation that cause the roots of it to be in the unit disk
From Simon Haykin's Adaptive Filter Theory: consider the characteristic equation is ${\displaystyle 1+a_1{z}^{-1}+a_2{z}^{-2}=0}$, then for the roots to be inside the unit circle (or in the unit disk), the coefficients of the quadratic must satisfy the following conditions: ${\displaystyle -1\le a_1+a_2,-1\le a_2-a_1}$, and ${\displaystyle -1\le a_2\le 1}$.
I could figure this out some of the way. For complex roots it follows that the coefficients lie in the region defined by the intersection of ${\displaystyle 4a_2>a_1^2}$ and ${\displaystyle a_2<1}$. For real roots, I could prove that ${\displaystyle \left|a_2\right|<1}$, and obviously ${\displaystyle 4a_2\le a_1^2}$. But I cannot figure out how the line boundaries of the region can be derived.

Rewrite the following exponential expressions as equivalent radical expressions. If the number is rational, write it without radicals or exponents.
${\displaystyle 2^{\frac{3}{2}}}$

Reduce quadratic form of 3 variables to sum of 3 squares
I am dealing with following quadratic form: ${\displaystyle q:{\mathbb{R}}^3\to \mathbb{R}}$
${\displaystyle q(x,y,z)=xy+yz+xz}$

Question regarding derivation of the quadratic formula
To derive the quadratic formula, I want to find (m,n) such that
${\displaystyle a{x}^2+bx+c+p={(mx+n)}^2+c}$
Expanding the RHS reveals
${\displaystyle a{x}^2+bx+c+p=m^2{x}^2+2mnx+n^2+c.}$
Then, ${\displaystyle a=m^2,b=2mn\Rightarrow (m,n)=(\pm \sqrt{a},\pm \frac{b}{2\sqrt{a}})}$
From here, isolating and solving for x would be trivial.Would this general method be correct? I'm confused as to the significance of p since ${\displaystyle p=n^2\Rightarrow n=\pm \sqrt{p}}$ here as well.
Perhaps it's been a long day or my algebraic skills have worsened, but I just couldn't seem to work out the quadratic formula with this path of thinking; there would always be some extra term...
I'm aware that dividing the entire original equation by a simplifies things a lot, but that should just be a specific case of this general form.

Question about changing a logarithm's base
I've been using the following method to derive/remember the logarithm base conversion formula: If I want to convert ${\displaystyle {\log }_a\left(x\right)}$ to an expression in base b, I say,
${\displaystyle a^{{\log }_a\left(x\right)}=x}$
${\displaystyle {\log }_b\left(a^{{\log }_a\left(x\right)}\right)={\log }_b\left(x\right)}$
${\displaystyle {\log }_a\left(x\right){\log }_b\left(a\right)={\log }_b\left(x\right)}$
${\displaystyle {\log }_a\left(x\right)=\frac{{\log }_b\left(x\right)}{{\log }_b\left(a\right)}}$
It kind of feels like I'm working backwards, and I was wondering if there's a more direct way to go about it. I tried:
${\displaystyle {\log }_a\left(x\right)={\log }_a\left(b^{{\log }_b\left(x\right)}\right)}$
${\displaystyle {\log }_a\left(x\right)={\log }_b\left(x\right){\log }_a\left(b\right)}$
but couldn't rid myself of the loga in the right-hand side of the equation. It occurred to me that if I could rewrite the "a" subscript as "${\displaystyle b^{\log b\left(a\right)}}$", I might be onto something. (Or might not.) Does the notation ever get used like that, where you perform a substitution in a subscript?

Quadratic equation with complex variable
Solve the following equation with ${\displaystyle \omega \in \mathbb{C}}$ and ${\displaystyle \gamma ,{\omega }_0\in \mathbb{R}}$ for its complex roots ${\displaystyle {\omega }_1,{\omega }_2}$
${\displaystyle {\omega }^2+\gamma \omega +{\omega }_0^2=0}$.

Quadratic formula and isolating f(x)
${\displaystyle f\left(x\right)-\frac{1}{f\left(x\right)}=g\left(x\right)}$

Prove that ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{n}{2n-1}=\frac{1}{2}}$