Solve Quadratic Equation: 2E(S+Wn)2

[fordy2707]

so I've been tasked with this insane question

obtain the solution of the following quadratic equation

View attachment 5476

how do I solve this, the 'E' (greek letter),is throwing me out,my complete hazardous guess is

2E(S+Wn)2 <--- 2 is a squared sorry my keyboard skills are lacking also 'tut'

any help greatly appreciated

 

[topsquark]

so I've been tasked with this insane question

obtain the solution of the following quadratic equation
how do I solve this, the 'E' (greek letter),is throwing me out,my complete hazardous guess is

2E(S+Wn)2 <--- 2 is a squared sorry my keyboard skills are lacking also 'tut'

any help greatly appreciated

\(\displaystyle s^2 + (2 \xi \omega _n)s + \omega _n^2 = 0\), I presume.

Use the quadratic formula. a = 1, \(\displaystyle b = 2 \xi \omega _n\) and \(\displaystyle c = \omega _n^2\).

-Dan

 

[I like Serena]

Hi fordy2707! Welcome to MHB! ;)

For the record, that greek 'E' is actually an 'x', or more specifically a 'xi' ($\xi$).
Anyway, as topsquark already suggested, just treat it as any other constant, say a $3$.

 

[I like Serena]

Oh, and if you have trouble writing that $\xi$, here's how it should be done:

;)

 

[fordy2707]

haha ,how are you getting those symbols up ??

 

[I like Serena]

haha ,how are you getting those symbols up ??

Type it as \$\xi\$.

Oh, and if you click Reply With Quote on topsquark's comment, you can see how the entire equation is formatted.

 

[fordy2707]

$\xi$ ${s}^{2}$

 

[fordy2707]

sorry if I am being totally stupid here (Doh) but is the answer to this quite literally

a = 1, \(\displaystyle b = 2 \xi \omega _n\) and \(\displaystyle c = \omega _n^2\)because i can't see how this can be calculated any further where I've just got symbols all over the place,

 

[topsquark]

sorry if I am being totally stupid here (Doh) but is the answer to this quite literally

a = 1, \(\displaystyle b = 2 \xi \omega _n\) and \(\displaystyle c = \omega _n^2\)because i can't see how this can be calculated any further where I've just got symbols all over the place,

It's a quadratic equation in s for all it's apparent complexity. \(\displaystyle \xi\) and \(\displaystyle \omega _n\) are just numbers. So
\(\displaystyle s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

If it's still too complicated looking switch out the \(\displaystyle \xi\) for z and \(\displaystyle \omega\) for w.

-Dan

- - - Updated - - -

Oh, and if you have trouble writing that $\xi$, here's how it should be done:

;)

i once had a Korean professor that called it "squiggly."

-Dan

 

[fordy2707]

once all is placed into the quadratic formula that's as far as I can go isn't it ?,as below that can't be calculated

So
\(\displaystyle s = \frac{-2\xi\omega_n \pm \sqrt{2\xi\omega n^{2} - 4s\omega_n^2}}{2s}\)

If it's still too complicated look 'Into giving up' I think is what one might say (Smirk)

 

[topsquark]

once all is placed into the quadratic formula that's as far as I can go isn't it ?,as below that can't be calculated

So
\(\displaystyle s = \frac{-2\xi\omega_n \pm \sqrt{2\xi\omega_n^{2} - 4s\omega_n^2}}{2s}\)

If it's still too complicated look 'Into giving up' I think is what one might say (Smirk)

a, b, and c are coefficients. s is the variable so you don't put that into the quadratic equation.

It looks like there is a common \(\displaystyle \omega _n^2\) in each term under the radical...

Oh, and if \(\displaystyle b = 2 \xi \omega _n\) what is b^2?

-Dan

 

[fordy2707]

\(\displaystyle s = \frac{-2\xi\omega_n \pm \sqrt{4\xi\omega_n - 4\omega_n^2}}{2}\)

now I can't go further can I as the figures are different ?

or would we be looking at

\(\displaystyle s = \frac{-2\xi\omega_n \pm \sqrt{\xi - 4\omega_n}}{2}\)

 

[topsquark]

\(\displaystyle s = \frac{-2\xi\omega_n \pm \sqrt{4\xi\omega_n - 4\omega_n^2}}{2}\)

now I can't go further can I as the figures are different ?

or would we be looking at

\(\displaystyle s = \frac{-2\xi\omega_n \pm \sqrt{\xi - 4\omega_n}}{2}\)

We have the quadratic equation \(\displaystyle s^2 + (2 \xi \omega _n )s + \omega _n ^2 = 0\)

So we know that \(\displaystyle a = 1\), \(\displaystyle b = 2 \xi \omega _n\), and \(\displaystyle c = \omega _n^2\). Thus
\(\displaystyle s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(\displaystyle s = \frac{-(2 \xi \omega _n) \pm \sqrt{(2 \xi \omega _n)^2 - 4(1)(\omega _n ^2)}}{2(1)}\)

\(\displaystyle s = \frac{-2 \xi \omega _n \pm \sqrt{4 \xi ^2 \omega _n^2 - 4 \omega _n^2}}{2}\)

\(\displaystyle s = \frac{-2 \xi \omega _n \pm \sqrt{(4 \omega _n^2) (\xi ^2 - 1)}}{2}\)

Can you finish from here?

It's pretty basic stuff for this level but I've known a lot of students who completely lose their heads when Greek letters come into it.

-Dan

 

[fordy2707]

apprehensively I have

\(\displaystyle s = \frac{-2 \xi \omega _n \pm (4 \omega _n) (\xi - 1)}{2}\)

 

[topsquark]

apprehensively I have

\(\displaystyle s = \frac{-2 \xi \omega _n \pm (4 \omega _n) (\xi - 1)}{2}\)

A lot of algebra mistakes in this one from start to finish. You may want to brush up on some of this.

\(\displaystyle \sqrt{(4 \omega _n^2) ( \xi ^2 - 1)} = \sqrt{4 \omega _n^2} \cdot \sqrt{\xi^2 - 1}\)

\(\displaystyle = 2 \omega _n \sqrt{ \xi ^2 - 1}\)

-Dan

 

[fordy2707]

I would not get this in a million years , it's annoying that I'm learning from a given 4 page teaching sheet explaining quadratic equations using factorisation which is easy,the examples all easy so I go to my supportive assignment and out the blue comes questions needing the quadratic formula which I had no prior explanation of so YouTube helped me learn that which was fine but then this hideous question comes up jeez how an Earth am I expected to know this from 4 pages of factorisation .i can see how your progressing, but for me to do it from the original equation myself is never going to happen.

 

[MarkFL]

I would not get this in a million years , it's annoying that I'm learning from a given 4 page teaching sheet explaining quadratic equations using factorisation which is easy,the examples all easy so I go to my supportive assignment and out the blue comes questions needing the quadratic formula which I had no prior explanation of so YouTube helped me learn that which was fine but then this hideous question comes up jeez how an Earth am I expected to know this from 4 pages of factorisation .i can see how your progressing, but for me to do it from the original equation myself is never going to happen.

If you were only exposed to factoring, but not completing the square and the resulting quadratic formula, then you were dealt an unreasonable problem. Typically, quadratic equation are taught in 3 sections. First, factoring and then completing the square, and finally you are shown that you can take the general quadratic, and complete the square on it to derive the quadratic formula. Each section should have lots of application problems, but one should never expect a student to work the problem you were given after only being exposed to factoring.

So please don't feel bad...this was not fair to you. (Shake)

 

[I like Serena]

I would not get this in a million years , it's annoying that I'm learning from a given 4 page teaching sheet explaining quadratic equations using factorisation which is easy,the examples all easy so I go to my supportive assignment and out the blue comes questions needing the quadratic formula which I had no prior explanation of so YouTube helped me learn that which was fine but then this hideous question comes up jeez how an Earth am I expected to know this from 4 pages of factorisation .i can see how your progressing, but for me to do it from the original equation myself is never going to happen.

There's a couple of power rules that I think should be part of anyone's repertoire.
$$\sqrt a = a^{1/2} \\
(ab)^c=a^c b^c \\
(a^b)^c=a^{bc} \\
a^{-c} = \frac{1}{a^c}
$$
Are you familiar with those? (Wondering)

In this particular case, we have:
$$\sqrt{4\omega_n^2 (\xi^2 -1)}
= \left(4\omega_n^2 (\xi^2 -1)\right)^{1/2} \\
= (4\omega_n^2)^{1/2}(\xi^2 -1)^{1/2} \\
= 4^{1/2}(\omega_n^2)^{1/2}\sqrt{\xi^2 -1} \\
= (2^2)^{1/2}(\omega_n^2)^{1/2}\sqrt{\xi^2 -1} \\
= 2^{2\cdot (1/2)}\omega_n^{2\cdot (1/2)}\sqrt{\xi^2 -1} \\
= 2^{1}\omega_n^{1}\sqrt{\xi^2 -1} \\
= 2\omega_n\sqrt{\xi^2 -1}
$$

 

[fordy2707]

So I have an update, chatted with my tutor today and he apologised after realising he had sent me the wrong quadratic equations supportive question papers to what I had been learning ,which was factoring simple questions ,although to my horror I still have to learn this in my next module .distance learning this is going to prove very difficult.

Thanks everyone for your input here anyway , and very soon I shall hassle you with more.

 

[topsquark]

So I have an update, chatted with my tutor today and he apologised after realising he had sent me the wrong quadratic equations supportive question papers to what I had been learning ,which was factoring simple questions ,although to my horror I still have to learn this in my next module .distance learning this is going to prove very difficult.

Thanks everyone for your input here anyway , and very soon I shall hassle you with more.

Hassle away! (Wave)

-Dan

 

[Monoxdifly]

For the record, that greek 'E' is actually an 'x', or more specifically a 'xi' ($\xi$).
Anyway, as topsquark already suggested, just treat it as any other constant, say a $3$.

Huh? I thought it was a 'Z' (zeta)...

 

[MarkFL]

Huh? I thought it was a 'Z' (zeta)...

xi: \(\displaystyle \xi\)

zeta: \(\displaystyle \zeta\)

 

[fordy2707]

so to revive this (dread) ,having not fully learned this properly yet but not wanting to walk away ,im getting my head around how you come to

$= 2\omega_n\sqrt{\xi^2 -1}$

but I am just wondering why can we not calculate and remove the square root of

$\sqrt{\xi^2-1}$

to become

$\sqrt{\xi-1}$ ?

is it because you can't square root the whole figure,the -1

 

[topsquark]

so to revive this (dread) ,having not fully learned this properly yet but not wanting to walk away ,im getting my head around how you come to

$= 2\omega_n\sqrt{\xi^2 -1}$

but I am just wondering why can we not calculate and remove the square root of

$\sqrt{\xi^2-1}$

to become

$\sqrt{\xi-1}$ ?

is it because you can't square root the whole figure,the -1

Just as \(\displaystyle (a+b)^2 \neq a^2 + b^2\) (You need to FOIL it out) so too \(\displaystyle \sqrt{a^2 + b^2} \neq a + b\)

-Dan