TS beat me to it. 11 to go! Your turn JohnInTX.

PLEASE! It's one thing to post this kind of stuff while the thread is dormant, but when the TS is engaged and participating, it really detracts from the conversation.

 

For ax^2+bx+c=0
x=(-b±√(b^2-4ac))/2a

a=1, b=-1,c=-1

x=(-*-1 ± √( -1^2 - 4*1*-1 )) /2*1
x=(1 ± √( 1 +4 ) /2
x=1 ± √5 /2
√5 /2 or √5???

 

Try
a=1
b=-2
c=-2

For your coefficients...

 

Don't write Req² as Req2. That appears either like a different quantity -- the resistance eq2 -- or as Req multiplied by two. The normal text expression is to use the carat symbol, '^', for exponentiation.

And, once again, start tracking your units!

(1 Ω)(Req) = (Req)(1 Ω) + (Req)^2 - (2 Ω)(1 Ω) - (2 Ω)(Req)
(Req)(1 Ω) + (Req)^2 - (2 Ω²) - (2 Ω)(Req) - (1 Ω)(Req) = 0
(Req)^2 + (Req)(1 Ω) - (2 Ω)(Req) - (1 Ω)(Req) - (2 Ω²) = 0
(Req)^2 + (Req)[1 Ω - 2 Ω - 1 Ω] - (2 Ω²) = 0
(Req)^2 - (2 Ω)(Req) - (2 Ω²) = 0

If I can be bothered to take the time to do it properly, why can't you?

Now apply the quadratic formula, ask yourself which of the two solutions makes sense, and you are done.

I think just because of your har work an interest clearly you were pilot of jet

 

For ax^2+bx+c=0
x=(-b±√(b^2-4ac))/2a
a=2 b=2 c=-2
x=(-2±√(2^2-4x2x-2))/2 x 2

for +ve
x=2.47213

This is where the sanity checks that you did (or were supposed to do) a long time ago come in.

Looking at the first three resistors, the middle one is in parallel with another resistance, so the extremes that are possible are that it is in parallel with an open and with a short. This places bounds of

2 Ω ≤ Req ≤ 3 Ω

So this part is consistent with your answer. But if we go one more level, then we see that the parallel resistance is also subject to these same bounds, which means

2 Ω + (1Ω || 2 Ω) ≤ Req ≤ 2 Ω + (1 Ω || 3 Ω)

This reduces to

2.67 Ω ≤ Req ≤ 2.75 Ω

Thus we already know what the value of Req must be to within 1.5% before we even begin. And we know that 2.47213 Ω is definitely NOT the correct answer.

So go back and ask yourself whether you chose a, b, and c correctly.

 

Try
a=1
b=-2
c=-2

For your coefficients...

ax^2- 2bx-2=0
so, a=1,b=-2, c=-2
For ax^2+bx+c=0

 

ax^2- 2bx-2=0
so, a=1,b=-2, c=-2
For ax^2+bx+c=0

Yes -- now solve it via the quadratic equation --- i.e. plug your formula into it...

 

How to solve ?
x=1 ± √5 /2
√(5 )/2 or √(5/2)

 

Write out the two expressions and then match up the coefficients

(Req)^2 - (2 Ω)(Req) - (2 Ω²) = 0
a(x^2) + b(x) + c = 0

Therefore,

a = 1
b = -2 Ω
c = -2 Ω²

 

How to solve ?
x=1 ± √5 /2
√(5 )/2 or √(5/2)

Walk through it step by step -- and show the steps here. You are SO close!

 

Your formula should be as follows:
Req^2-2*Req-2Ω=0

 

I think just because of your har work an interest clearly you were pilot of jet

Attention to detail is very, vary valuable for a pilot, but probably it is even more valuable to the scientist and engineer.

 

Write out the two expressions and then match up the coefficients

(Req)^2 - (2 Ω)(Req) - (2 Ω²) = 0
a(x^2) + b(x) + c = 0

Therefore,

a = 1
b = -2 Ω
c = -2 Ω²

how you got that Ω²

 

Attention to detail is very, vary valuable for a pilot, but probably it is even more valuable to the scientist and engineer.

How to get it?
is there any test for it.

 

how you got that Ω²

I think that was a typo? --- please see post #511

 

how you got that Ω²

The superscript characters used to be in the symbol table (back on vBulletin) but they aren't anymore.

But you can access them via the Alt-number shortcuts. In this case, it's Alt-253.

One of the stickies has tables of the Alt-number characters. Very handy. The few I use I have memorized. Others I can usually find pretty quickly by trial and error. Others I have to break down and go look at the sticky.

 

√(5 )/2 or √(5/2)
Walk through it step by step -- and show the steps here. You are SO close!

I mean to say will it be √(5 )/2 = √2.5
or
√(5/2)=2.2/2?

 

sorry, i have to go for litle bit.
Actually i am going in university to post my resume.

 

I think that was a typo? --- please see post #511

Since the first term is (Req)^2, each term has to have units of resistance-squared. The coefficient of the linear term therefore has to have units of resistance and the constant term has to have units of resistance-squared. Note that the square on the constant term only applies to the unit, just as ab^2 is not the same as (ab)^2.

 

If we're to help find the error we need to see your work

For starters you should have
Req^2-2*Req-2=0

--Amended for consistency--