Master Algebra 1 Word Problems with our Comprehensive Practice

A continuous-time Markov chain with state space S nonzero transition rates q(1, 2) = g(2, 3) = q(3, 1) = q(4, 1) = 1 and q(1,4) = q(3, 2) = q(3, 4) = g(4, 3) = 2.
(a) Provide (i) the transition rate graph, (ii) the holding time parameters, (iii) the generator matrix, and (iv) the transition matrix for the embedded Markov chain.
(b) How long on average will the Markov chain stay in each state before moving to the next state?
(c) If the chain is at state 3 and moves next to state 4, how long on average will it stay in state 3 in this case?

Three integers are in the ratio 2:3:8. If 4 is added to the middle number, the resulting number is the second term of a geometric progression of which the other two integers are the first and third terms. How do you find the three integers?

Find d for the arithmetic series with s17=-170 and a1=2

Write an equation of a line passing through (4, -3), perpendicular to y=−4x+3

Find the next two terms of the geometric sequence ${\displaystyle \frac{1}{3},\frac{5}{6},\frac{25}{12},...}$

Write an equation in point-slope form for the given (−2, −7) m = 1

I've got a problem with factorising the denominator of an LTI system. The system is a simple boost converter and my current transfer function is
$$U_O=\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }\cdot <!--\; \cdot \; -->U_I-<!--\; -\; -->I_O\cdot <!--\; \cdot \; -->L\cdot <!--\; \cdot \; -->s}{C\cdot <!--\; \cdot \; -->L\cdot <!--\; \cdot \; -->s^2+{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime 2}}.$$
I want to solve for the systems poles next and and bring them into time constant representation. I thus use the common approach
$$C\cdot <!--\; \cdot \; -->L\cdot <!--\; \cdot \; -->s^2+{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime 2}\stackrel{!}{=}0$$
which can be solved with the quadratic formula
$$s=\pm <!--\; \pm \; -->\frac{\sqrt{-<!--\; -\; -->4\cdot <!--\; \cdot \; -->C\cdot <!--\; \cdot \; -->L\cdot <!--\; \cdot \; -->{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime 2}}}{2\cdot <!--\; \cdot \; -->C\cdot <!--\; \cdot \; -->L}=\pm <!--\; \pm \; -->\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }}{\underset{T_{CL}}{\underset{⏟<!--\; ⏟\; -->}{\sqrt{C\cdot <!--\; \cdot \; -->L}}}}j=\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }}{T_{CL}}j$$
and leads to the factorised form
$$(s+\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }}{T_{CL}}j)\cdot <!--\; \cdot \; -->(s-<!--\; -\; -->\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }}{T_{CL}}j).$$
This result is obviously wrong since expanding the term doesn't give the same result and I'm not able to wrap my head around why this is the case.
Factorising the denominator with the third binomial formula gives the correct factors instead
$$(T_{CL}\cdot <!--\; \cdot \; -->s+{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }j)\cdot <!--\; \cdot \; -->(T_{CL}\cdot <!--\; \cdot \; -->s-<!--\; -\; -->{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }j)$$
and these are already in time constant representation, which leads to the fully factorised transfer function in time constant representation
$$U_O=\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }\cdot <!--\; \cdot \; -->U_I-<!--\; -\; -->I_O\cdot <!--\; \cdot \; -->L\cdot <!--\; \cdot \; -->s}{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime 2}\cdot <!--\; \cdot \; -->\underset{p}{\underset{⏟<!--\; ⏟\; -->}{(\frac{T_{CL}}{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }}\cdot <!--\; \cdot \; -->s+j)}}\cdot <!--\; \cdot \; -->\underset{p^{\ast <!--\; \ast \; -->}}{\underset{⏟<!--\; ⏟\; -->}{(\frac{T_{CL}}{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }}\cdot <!--\; \cdot \; -->s-<!--\; -\; -->j)}}}.$$
Where did I mess up the quadratic formula?
My only approach is that the coefficients aren't
$$\begin{array}{rl}a& =C\cdot <!--\; \cdot \; -->L\\ b& =0\\ c& ={\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime 2}\end{array}$$
and I'm not solving for s, but that the coefficients are
$$\begin{array}{rl}a& =1\\ b& =0\\ c& ={\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime 2}\end{array}$$
and I'm solving for $T_{CL}\cdot <!--\; \cdot \; -->s$ with respect to the quadratic equation
$${\underset{x}{\underset{⏟<!--\; ⏟\; -->}{(T_{CL}\cdot <!--\; \cdot \; -->s)}}}^2+{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime 2}.$$
Please give some advice on how to deal with such situations, I'm seriously confused and couldn't find an answer browsing through the material about quadratic equations. Different problems are far to over represented.

Write an equation of a line through (-3,1) and perpendicular to y= -3x+7

Can anyone invert
$y=a\cos <!--\; \; -->(x)+b\sin <!--\; \; -->(2x)$
to give $x=f(y)$?

Find the slope of a line that is perpendicular to a slope of -1/6

A geometric sequence is defined recursively by ${\displaystyle a_n=4a_{n-1}}$, and the first term of the sequence is 0.5, how do you find the explicit formula?

Find the sum of the arithmetic series 1 + 3 + 5 + ... + 27

Finding the range of $\frac{5\cos <!--\; \; -->x-<!--\; -\; -->2{\sin }^2<!--\; \; -->x+4\sin <!--\; \; -->x-<!--\; -\; -->3}{6|\cos <!--\; \; -->x|+1}$

The sum of 3 times Darlene's age and 7 times Sharon's is 173. Darlene is 2 years less than twice as old as Sharon is. How do you find each of their ages?

How do you write the first five terms of the sequence defined recursively ${\displaystyle a_1=81,a_{k+1}=\frac{1}{3}{a}_k}$, then how do you write the nth term of the sequence as a function of n?

Find the slope of any line perpendicular to the line passing through (−7,3) and (−14,14)

Find the number of terms in the following geometric series: 1 + 2 + 4 + ... + 67108864

Find the slope perpendicular to the line y=10

Find the equation of the line that passes through (−1,1) and is perpendicular to the line that passes through the following points: (−2,4),(−12,21)