How to find the values of x in the equation (1)?

[mspaic]

ok this is a question in some algebra assignment.

it's drivin' me kinda crazy and nobody will even help in university. this should be easy to solve for some of you i know.

keep in mind that K1 K2 and K3 and M1 and M2 are all constants that are always positive. K stands for constant of as spring and M is the mass i belive, so they are all positive values.

here is the question:

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SHOW THAT X1 AND X2 (two possible values of x to solve the equation) ARE BOTH DISTINCT AND NEGATIVE:

( X + (K1 +K2)/M1 ) * ( x + (K2 + K3)/m2) - ( -K2/M1) * (-K2/M2) =0


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i simplified it (assumin i made no mistakes) and got :

x^2 (M1)(M2) + x ( (K1+K3)(M1) + (K1+K2)(M2) ) + (K1)(K2)+ (K1)(K3) +(K2)(K3) =0

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well since all the m's and k's are positive, it's easy for me to prove to him that the x values have to always be negative. i can do this with a statement or whatever. BUT, i'd like to find the values of x (there are two and they are both negative) because it will help me for the later questions. so if anyone knows how to do that, i'd greatly appreciate it.

 

[mspaic]

does anyone know it..

or is it even possible

the teacher assistant i think hinted that it was possible to find the two values of x

 

[Justin Lazear]

If you have a quadratic form, why don't you just use the quadratic equation to solve for the roots?

--Justin

 

[mspaic]

is it possible to do a quadratic formula with these values
like i don't have values for the masses or the k's .
i will try it

 

[mspaic]

i just put it into the quadratic and it ain't going to work errrr like the stuff underneath the square root gets way too complicated

 

[Justin Lazear]

Nobody ever said it was going to be pretty.

And why wouldn't you be able to use the quadratic formula?

--Justin

 

[mspaic]

but underneath the root you have like terms you can't simplify
and for x I'm supposed to get a number ...

 

[BinaryFinary]

How are you supposed to get a number for x when all you have are variables?

 

[Justin Lazear]

You can only find the roots in terms of m1,m2,k1,k2,k3. You cannot find numerical roots unless you have numerical values of the above constants.

Keep in mind that an expression containing m1,m2,k1,k2,k3 will be a number once you substitute in values for m1,m2,k1,k2,k3.

--Justin

 

[mspaic]

ehe ok finally

i'll just explain to him why x will always be negative though

 

[Leong]

i simplify the equation and get
$$b^2-4ac=(m_{1}m_{2})^2\{[m_{1}(k_{2}+k_{3})-m_{2}(k_{1}+k_{2})]^2+4k^2\}(>0)$$
So, x1 and x2 are distinct but i can't prove they are negative.

 

[mspaic]

x^2 (M1)(M2) + x ( (K1+K3)(M1) + (K1+K2)(M2) ) + (K1)(K2)+ (K1)(K3) +(K2)(K3) =0

look at this

you know the first and the last term will be positive right since k and m are always positive

therefore, the middle term has to be negative, for the equation to =0
thus, x has to be negative

 

[Leong]

i think you have made a mistake in your simplification.
$$(m_{1}m_{2})^{2}x^2+m_{1}m_{2}[m_{1}(k_{2}+k_{3})+m_{2}(k_{1}+k_{2})]x+m_{1}m_{2}(k_{1}+k_{2})(k_{2}+k_{3})-k_{2}^2=0$$

 

[mspaic]

the term with m1m2(k1 +k2)(k2+k3)

why would u need the mi and m2

when you multiply you alread have them on the bottom so when you common factor why do u need it up there

 

[mspaic]

can someone else factor it please see if i have it right it shouldn't take u long

 

[maverick280857]

Leong has proved that the roots are real and distinct (using the discriminant condition). I haven't tried this but I think you should be able to prove that the roots are negative by proving that the product of the roots is negative, i.e the constant term and the coefficient of the second degree term are of opposite signs.

 

[mspaic]

i simplify the equation and get
$$b^2-4ac=(m_{1}m_{2})^2\{[m_{1}(k_{2}+k_{3})-m_{2}(k_{1}+k_{2})]^2+4k^2\}(>0)$$
So, x1 and x2 are distinct but i can't prove they are negative.

ok I'm pretty sure i figured out how to prove that they are negative.

however, can someone please do similar to what he did above (just for my simplification) to prove that the roots are distinct. I'm sure my simplification is right because i confirmed it with numerous classmates.

 

[mspaic]

x^2 (M1)(M2) + x ( (K1+K3)(M1) + (K1+K2)(M2) ) + (K1)(K2)+ (K1)(K3) +(K2)(K3) =0

there is my simplification by the way

 

[mspaic]

anyone...

 

[Leong]

Simplification

$$[x+\frac{k_{1}+k_{2}}{m_1}]*[x+\frac{k_{2}+k_{3}}{m_2}]-(\frac{-k_{2}}{m_1})*(\frac{-k_{2}}{m_2})=0...(1)$$
$$(1)*m_{1}m_{2}:$$
$$[xm_{1}m_{2}+(k_{1}+k_{2})m_{2}]*[xm_{1}m_{2}+(k_{2}+k_{3})m_{1}]-k_{2}^2=0$$
$$(m_{1}m_{2})^2x^2+x[m_{1}^2m_{2}(k_{2}+k_{3})+m_{1}m_{2}^2(k_{1}+k_{2})]+m_{1}m_{2}(k_{1}+k_{2})(k_{2}+k_{3})-k_{2}^2=0$$
$$(m_{1}m_{2})^2x^2+xm_{1}m_{2}[m_{1}(k_{2}+k_{3})+m_{2}(k_{1}+k_{2})]+m_{1}m_{2}(k_{1}+k_{2})(k_{2}+k_{3})-k_{2}^2=0$$

 

[Leong]

Discriminant

$$b^2-4ac=(m_{1}m_{2})^2[m_{1}(k_{2}+k_{3})+m_{2}(k_{1}+k_{2})]^2-4(m_{1}m_{2})^2[m_{1}m_{2}(k_{1}+k_{2})(k_{2}+k_{3})-k_{2}^2]$$
$$=(m_{1}m_{2})^2\{[m_{1}(k_{2}+k_{3})]^2+2m_{1}m_{2}(k_{2}+k_{3})(k_{1}+k_{2})+[m_{2}(k_{1}+k_{2})]^2-4m_{1}m_{2}(k_{1}+k_{2})(k_{2}+k_{3})+4k_{2}^2\}$$
$$=(m_{1}m_{2})^2\{[m_{1}(k_{2}+k_{3})]^2-2m_{1}m_{2}(k_{2}+k_{3})(k_{1}+k_{2})+[m_{2}(k_{1}+k_{2})]^2+4k_{2}^2\}$$
$$=(m_{1}m_{2})^2\{[m_{1}(k_{2}+k_{3})-m_{2}(k_{1}+k_{2})]^2+4k_{2}^2\}(>0)$$

 

[mspaic]

thanks a lot man

 

[Leong]

Correction : Simplification & Discriminant

What I have written above are totally wrong. I made mistakes.
$$[x+\frac{k_{1}+k_{2}}{m_1}]*[x+\frac{k_{2}+k_{3}}{m_2}]-(\frac{-k_{2}}{m_1})*(\frac{-k_{2}}{m_2})=0...(1)$$
$$(1)*m_{1}m_{2}:$$
$$[m_{1}x+(k_{1}+k_{2})][(m_{2}x+(k_{2}+k_{3})]-k_{2}^2=0$$
$$m_{1}m_{2}x^2+x[m_{2}(k_{1}+k_{2})+m_{1}(k_{2}+k_{3})]-k_{2}^2=0$$
$$b^2-4ac=[m_{2}(k_{1}+k_{2})+m_{1}(k_{2}+k_{3})]^2-4m_{1}m_{2}(-k_{2}^2)$$
$$=[m_{2}(k_{1}+k_{2})+m_{1}(k_{2}+k_{3})]^2+4m_{1}m_{2}k_{2}^2(>0)$$
Since
$$\sqrt{b^2-4ac}>b$$,then
$$-b+\sqrt{b^2-4ac}>0 \ and \ -b-\sqrt{b^2-4ac}<0$$
So, there is one positive and one negative roots.