- Thread starter vec1
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Use the quadratic formula, after putting the equation into standard form

Ax^2 + Bx + C = 0

As a check, the discriminant (B^2 - 4AC) will be N^2 - 4

(That is, the solutions are Real numbers when N is 2 or more in absolute value. Otherwise, the solutions are Complex with an imaginary part.)

Edited. Thank you, Jomo.

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And NAdding to MarkFL's hint:

Use the quadratic formula, after putting the equation into standard form

Ax^2 + Bx + C = 0

As a check, the discriminant (B^2 - 4AC) will be N^2 - 4

(That is, the solutions are Real numbers when N is 2 or more. Otherwise, the solutions are Complex with an imaginary part.)

Lets say that N = 3.33333 so x = 3 for this example. So I tried to put this into quadratic form and get this

x^2 - Nx = -1 and x^2 - Nx +1 = 0

so it kinda looks like the quadratic form but not quite . Any ideas?

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Yes, that's the correct quadratic in \(x\):

Lets say that N = 3.33333 so x = 3 for this example. So I tried to put this into quadratic form and get this

x^2 - Nx = -1 and x^2 - Nx +1 = 0

so it kinda looks like the quadratic form but not quite . Any ideas?

\(\displaystyle x^2-Nx+1=0\)

Now, to solve for \(x\), either complete the square or apply the quadratic formula.

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Not so, I agreed with Otis that N

Lets say that N = 3.33333 so x = 3 for this example. So I tried to put this into quadratic form and get this

x^2 - Nx = -1 and x^2 - Nx +1 = 0

so it kinda looks like the quadratic form but not quite . Any ideas?

If you thought that I said

You say

So x + 1/x =N, ie x^2 -Nx + 1 =0. Now N=3.33333. So x^2 -3.33333x + 1= 0.

1st of all x= 3 is not a solution to this equation. Note that 3^2 -3.33333*3 + 1 = 9 - 9.99999 + 1 \(\displaystyle \neq\) 0.

In this quadratic you have a =1, b= -3.33333 and c=1

So x= ....

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Gah! There are too many decimals here for a Math post. We're talking about x = 3 and N = 1/3. Math deals with exact values when it can, not decimal approximations to the numbers.Jomo says that N must be equal or less than -2 for this to work.

Lets say that N = 3.33333 so x = 3 for this example.

-Dan

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I'm guessing that you actually started with x=3 and found \(\displaystyle N = 3 + \frac{1}{3} = 3 \frac{1}{3} = \frac{10}{3} \approx 3.33333\). So when you use the quadratic formula, you'll get onlyLets say that N = 3.33333 so x = 3 for this example. So I tried to put this into quadratic form and get this

x^2 - Nx = -1 and x^2 - Nx +1 = 0

so it kinda looks like the quadratic form but not quite . Any ideas?

When you say it isn't quite "quadratic form", are you saying you aren't used to seeing a parameter like N in a quadratic equation? You'll be seeing this a lot eventually; it definitely

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Dan 10/3 ~ 3.33333, not 1/3Gah! There are too many decimals here for a Math post. We're talking about x = 3 and N = 1/3. Math deals with exact values when it can, not decimal approximations to the numbers.

-Dan

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No worries. I'm a Physicist and it's only off by an order of magnitude. Good enough!Dan 10/3 ~ 3.33333, not 1/3

Thanks for the catch!

-Dan

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For what to work, vec1?Jomo says that N must be [less than or equal to] -2 for this to work ...

We can solve the given equation for x, without knowing a value for N. In other words, it doesn't matter what N is. Simply treat N as a constant.

The commentary about |N| >= 2 is only an aside because you didn't specify whether the solutions must be Real or not. (At this point, I regret posting about how N's value affects the nature of the solutions for x because that commentary seems to have sent you off-track.)

We want to solve x^2 - Nx + 1 = 0 for x.

That equation is in standard form, so we can see that the coefficients are:

A = 1

B = -N

C = 1

Therefore, the quadratic formula gives the solutions as:

x = N/2 + sqrt(N^2 - 4)/2

or

x = N/2 - sqrt(N^2 - 4)/2

Again, in this exercise, it doesn't matter what N is. I had given you the discriminant as a check for your own set-up. Cheers

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That is one of the two solutions, and it comes straight from using the quadratic formula (shown below in red).... I want to find out how x = N/2 - sqrt(N^2 - 4)/2 is derived

Given a quadratic equation in standard form:

Ax^2 + Bx + C = 0

the two solutions for x are written in terms of the coefficients A,B,C:

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

or

x = -B/(2A) - sqrt(B^2 - 4AC)/(2A)

Both solutions are required, and note the difference between the two.

In your exercise, you wrote the quadratic equation:

x^2 - N*x + 1 = 0

This is in standard form, so we can see that:

A = 1

B = -N

C = 1

Substitute those expressions for A,B,C into the quadratic formula and simplify.

If you would like more help with this exercise, please show your work. Cheers

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Ax^2 + Bx + C = 0

and complete the square.

Alternatively, you can google keywords derive quadratic formula and watch a video of the process.

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Are you saying that you do not know how a quadratic equation is solved?I should have wrote 3 1/3 instead of 3.333333

so i will go over x^2 - Nx + 1 = 0 again and work out the details on that

I want to find out how x = N/2 - sqrt(N^2 - 4)/2 is derived

\(\displaystyle \begin{align*}px^2+qx+t&=0,~p> 0 \\x^2+\left(\frac{q}{p}\right)x&=\frac{-t}{p}\\\left(x+\frac{q}{2p}\right)^2&=\frac{q^2}{4p^2}-\frac{t}{p} \\x+\frac{q}{2p}&=\frac{\pm\sqrt{q^2-4pt}}{2p}\\x&=-\frac{q}{2p}+\frac{\pm\sqrt{q^2-4pt}}{2p}\end{align*}\)

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Dan, seriously can you please explain to me why being off by any power or 10 is ok for a physicist. The first time I ran across this I was asking my physics professor to find a mistake in my work. He lookrd and said it was correct. After I pointed out that the answer was wrong and was off by a decimal place he through me out his office!No worries. I'm a Physicist and it's only off by an order of magnitude. Good enough!

Thanks for the catch!

-Dan

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Jomo, my dear boy, it was a joke!Dan, seriously can you please explain to me why being off by any power or 10 is ok for a physicist. The first time I ran across this I was asking my physics professor to find a mistake in my work. He lookrd and said it was correct. After I pointed out that the answer was wrong and was off by a decimal place he through me out his office!

Seriously, though, yes Physicists do tend to accept rather blatent errors in their data. It all depends on the equipment and the data. If the statistical error is large enough then yes, we can tolerate an answer within about an order of magnitude. However, I don't accept that... but then I'm just a dumb theorist.

-Dan

Addendum: I was taking a graduate Physics lab and had the task of looking over some interactions involving kaons (I think) on a projection of a bubble chamber. I had to track down a specific interaction. After I had turned the lab in (very unsatisfied with it) my professor lauded me for discovering five new particles. Yes, my ability to take data is that bad.