Quadratic Equation help?

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vineo76
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Quadratic Equation help?

Post by vineo76 » Tue Feb 10, 2009 9:09 pm

Can someone show me step by step how to put this in quadratic form?
its specifaclly for JFET design by the way (Common Source Amplifier)


A(v) = -(20k||R(d) / 3k-R(d)+350=-4


where k is for kilo ohms..

how do I put it in quadratic form?

Can you show me step by step?

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jwax
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Re: Quadratic Equation help?

Post by jwax » Wed Feb 11, 2009 2:46 am

There's better math wizards on this forum than I, but that looks like a poorly-written linear equation.
What is the || symbol?

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MrAl
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Re: Quadratic Equation help?

Post by MrAl » Wed Feb 11, 2009 4:37 am

Hi there,


John:
I assume those two lines || mean "in parallel with". In my reply here i use the
functional notation "para(R1,R2)" to indicated R1 and R2 are in parallel and is
equal to this:
para(R1,R2)=R1*R2/(R1+R2)
so that means
R1||R2=para(R1,R2)=R1*R2/(R1+R2)


vineo:
vineo76 wrote:Can someone show me step by step how to put this in quadratic form?
its specifaclly for JFET design by the way (Common Source Amplifier)
A(v) = -(20k||R(d) / 3k-R(d)+350=-4
Because your equation is not syntaxically correct, we can assume a few different forms for your equation.

If we assume you meant this:
-para(20000,Rd)/3000-Rd+350=-4

then we get this:
-(20*Rd)/(3*(Rd+20000))-Rd+350=-4

which when simplified we get this:
-3*Rd^2-58970*Rd+21000000=-12*Rd-240000

and then subtracting -12*Rd-240000 from both sides we get this:
-3*Rd^2-58958*Rd+21240000=0

and that is in quadratic form, and the two possible solutions are:
Rd=-(sqrt(932731441)+29479)/3
Rd=(sqrt(932731441)-29479)/3

The second one here leads to a positive value for Rd so we assume
that is the right one, and we get
Rd=353.88 ohms.

If instead we assume you meant this:
-para(20000,Rd)/(3000-Rd+350)=-4

then we get this quadratic:
-4*Rd^2-86600*Rd+268000000=0

the two possibile solution are:
Rd=-25*sqrt(294689)-10825
Rd=25*sqrt(294689)-10825

and again the second one is the right one, which gives us
Rd=2746 ohms.

If instead you meant this:
-para(20000,Rd)/(3000-Rd)+350=-4;

then we get this quadratic:
354*Rd^2+6038000*Rd-21240000000=0

the two possible solutions are:
Rd=-(500*sqrt(16633321)+1509500)/177
Rd=(500*sqrt(16633321)-1509500)/177

and again the second one is the right one, which gives us
Rd=2993 ohms
LEDs vs Bulbs, LEDs are winning.

vineo76
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Re: Quadratic Equation help?

Post by vineo76 » Wed Feb 11, 2009 4:24 pm

If instead we assume you meant this:
-para(20000,Rd)/(3000-Rd+350)=-4

then we get this quadratic:
-4*Rd^2-86600*Rd+268000000=0

the two possibile solution are:
Rd=-25*sqrt(294689)-10825
Rd=25*sqrt(294689)-10825

and again the second one is the right one, which gives us
Rd=2746 ohms.


I meant the one above..and I knew the answer. I have ti, but when I write it out, I wonder how the author got there. Is there a certain formula to use? Or do I have to write it all out? How did you get your answer?

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MrAl
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Re: Quadratic Equation help?

Post by MrAl » Wed Feb 11, 2009 4:38 pm

vineo76 wrote:If instead we assume you meant this:
-para(20000,Rd)/(3000-Rd+350)=-4

then we get this quadratic:
-4*Rd^2-86600*Rd+268000000=0

the two possibile solution are:
Rd=-25*sqrt(294689)-10825
Rd=25*sqrt(294689)-10825

and again the second one is the right one, which gives us
Rd=2746 ohms.


I meant the one above..and I knew the answer. I have ti, but when I write it out, I wonder how the author got there. Is there a certain formula to use? Or do I have to write it all out? How did you get your answer?

Hi again,


We start with this equation which is the one you gave:

-para(20000,Rd)/(3000-Rd+350)=-4

We then calculate the parallel resistance as shown, simplify, and subtract until
we get a zero (0) on the right side like so:

-4*Rd^2-86600*Rd+268000000=0

and now this form is said to be "in quadratic form" because it is in the same form as
this general quadratic:

A*x^2+B*x+C=0

if we set:
A=-4
B=-86600
C=268000000
and we call the variable to be solved for "Rd" instead of "x".

Once we have the equation in quadratic form we can then apply what is called the
"Quadratic Formula" which looks like this:

x= (-B+sqrt(B^2-4*A*C))/(2*A)
x= (-B-sqrt(B^2-4*A*C))/(2*A)

which as you can see gives two values for x (the first equation and the second equation).
We then compute both of these to see which value x should become.
In cases where it is a resistor, it should come out to a positive value.
In cases where there are two positive values, there could be two distinct solutions both of
which should be tried in the application to see which one fits best.
There are also cases where you get two imaginary solutions when the part called the
"discriminant" (the part inside the radical: "B^2-4*A*C") is negative, which could mean
there are no solutions, or there are two imaginary solutions who's interpretation depends
on the actual application this formula is being used for.


You could look on the web if you want to find out how the Quadratic Formula is derived
or simply use it as shown.
LEDs vs Bulbs, LEDs are winning.

Bigglez
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Re: Quadratic Equation help?

Post by Bigglez » Wed Feb 11, 2009 10:48 pm

MrAl wrote: and now this form is said to be "in quadratic form" because it is in the same form as
this general quadratic:

A*x^2+B*x+C=0
..
You could look on the web if you want to find out how the Quadratic Formula is derived
or simply use it as shown.
Well played, sir! QED

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