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?

## Quadratic Equation help?

### Re: Quadratic Equation help?

There's better math wizards on this forum than I, but that looks like a poorly-written linear equation.

What is the || symbol?

What is the || symbol?

### Re: Quadratic Equation help?

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:

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

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:

Because your equation is not syntaxically correct, we can assume a few different forms for your equation.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

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.

### Re: Quadratic Equation help?

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?

-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?

### Re: Quadratic Equation help?

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.

### Re: Quadratic Equation help?

Well played, sir! QEDMrAl 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.

### Who is online

Users browsing this forum: Bing [Bot] and 20 guests