Thread: determine values of ω, that satisfy the roots given.....

i have an assignment question hopefully someone can help me with or point me in the right direction....

the question comes in 3 parts.i have done parts (i),(ii) i need to work out the 3rd part(iii).

(ii) if the roots of a cubic equation are x= -5/2 , x= 1/4, x= 3
determine the cubic expression by expansion, with these roots.


(iii) if x = Cos(2ω) in part (ii) of this question where ω is measured in radians. determine the values of ω, that satisfy the roots given that
0 ≤ ω ≥ 2(pi)


any help much appreciated

*

Hello, decoy808!

The question comes in 3 parts. I have done parts (i), (ii).

(ii) If the roots of a cubic equation are: $\displaystyle x\:=\:\text{-}\tfrac{5}{2},\:\tfrac{1}{4},\:3$
determine the cubic expression, by expansion, with these roots.

We have: .$\displaystyle \begin{array}{c}x \:=\:\text{-}\frac{5}{2} \\ \\[-4mm] x \:=\:\frac{1}{4} \\ \\[-4mm] x \:=\:3 \end{array}\quad\Rightarrow\quad \begin{array}{cc} 2x+5 \:=\:0 &{\color{blue}[1]}\\ 4x-1 \:=\:0 &{\color{blue}[2]}\\ x-3\:=\:0 & {\color{blue}[3]} \end{array}
$

Multiply [1]. [2] and [3]: .$\displaystyle f(x) \;=\;(2x+5)(4x-1)(x-3)$


Therefore: . $\displaystyle f(x) \;=\;8x^3 - 6x^2 - 59x + 15$

(iii) $\displaystyle x = \cos2\theta$ in part (ii) of this question, where $\displaystyle \theta$ is measured in radians.
Determine the values of $\displaystyle \theta$ that satisfy the roots, given that $\displaystyle 0 \leq \theta \leq 2\pi$

We have: .$\displaystyle \left(2\cos2\theta + 5\right)\left(\cos2\theta - 3\right)\left(4\cos2\theta - 1\right) \:=\:0$


And we have three equations to solve:

. . $\displaystyle 2\cos\theta + 5 \:=\:0\quad\Rightarrow\quad \cos2\theta \:=\:\text{-}\tfrac{5}{2} \quad\hdots\quad\text{ No real roots.}$

. . $\displaystyle \cos2\theta - 3 \:=\:0 \quad\Rightarrow\quad \cos2\theta \:=\:3 \quad\hdots\quad\text{ No real roots.}$


. . $\displaystyle 4\cos2\theta - 1 \:=\:0 \quad\Rightarrow\quad \cos2\theta \:=\:\tfrac{1}{4}$

. . . . $\displaystyle 2\theta \;\approx\;\begin{Bmatrix}1.318116 \\ 4.965069 \\ 7.601301 \\ 11.248255 \end{Bmatrix} \quad\Rightarrow\quad \theta \;\approx\;\begin{Bmatrix}0.654 \\ 2.483 \\ 3.801 \\ 5.624 \end{Bmatrix}$

thanks soroban it is a great help. i am still trying to understand a few things please excuse my ignorance

why did u subsitute (omega)for (theta)?

Hello, decoy808!

why did u subsitute (theta)for (omega)?

I's just easier to type.

My fingers are used to typing \-t-h-e-t-a,
. . while I have spell \-o-m-e-g-a very carefully.