For the discussion of math. Duh.

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Lyon
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Ok...so last sunday I was really bored, and for absolutely no reason in particular I started to play around with the quadratic formula. I ended up coming up with 3 different equations...for a, for b, and c. Unfortunately, the equations every other variable; for example to find a, you need b, c, and x. Even worse, to find c, you need a, b, x, AND c...so if you have c, you can find it! I tested them out with some standard quadratic equations...so here you go. If anybody can find a practical use for this, please share it with me, because I NEED to find some way to justify it to myself!

A = ( ( (bx+2c)/-x)^2 - c^2)/-4c

C = square root of ((-2(bx+c)+bx)/x)^2+4ac (the whole thing is under the square root...I don't know a symbol for square root unfortunately

B = square root of (the whole thing is under the square root again) (2a^2x^2+2abx+2ac+b)^2 + 4ac

Dopefish
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I take it you just rearranged the quadratic formula? I haven't looked too closely, but hopefully you dealt with the plus or minuis bits appropriately (although you probably squared that part so it'd be ok).

Anyhow... a, b, and c are typically known, so formulas for them strike me as unlikely to be useful. Espcially in the case of b and c, where you didn't actually isolate them.

skullturf
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Wouldn't the formula for c be

c = - ax^2 - bx

?

mdyrud
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Even though the result is a unwieldy and probably not terribly useful, good job. One of the best ways to get better at math is exposing yourself to it, and it sounds like this is a really good way to give some great practice with rearranging equations. So don't worry about justifying yourself. Embrace your inner geek. After you've gotten bored of doing that, move on to proofs. The Fibonacci Sequence has an even number every third term? Not exactly a useful fact to know, but it's something to prove, and it gives you practice that will be useful in other, more practical problems.

Mike_Bson
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mdyrud wrote:Even though the result is a unwieldy and probably not terribly useful, good job. One of the best ways to get better at math is exposing yourself to it, and it sounds like this is a really good way to give some great practice with rearranging equations. So don't worry about justifying yourself. Embrace your inner geek. After you've gotten bored of doing that, move on to proofs. The Fibonacci Sequence has an even number every third term? Not exactly a useful fact to know, but it's something to prove, and it gives you practice that will be useful in other, more practical problems.

This. I always find myself messing with fun stuff like this.

thicknavyrain
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mdyrud wrote:The Fibonacci Sequence has an even number every third term?

Sonofa...
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undecim
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If you start with ax^2 + bx + c = 0, it's pretty simple.

a = -(bx + c)/x^2
b = -(ax^2+c)/x
c = -(ax^2 + c)

Though I suppose starting with the quadratic formula is more fun.
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Mike_Bson
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thicknavyrain wrote:
mdyrud wrote:The Fibonacci Sequence has an even number every third term?

Sonofa...

Also, if a Fibonacci number is the nth number, n is prime. For example, 89 is prime, and the 11th Fibonacci number.

thicknavyrain
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Mike_Bson wrote:
thicknavyrain wrote:
mdyrud wrote:The Fibonacci Sequence has an even number every third term?

Sonofa...

Also, if a Fibonacci number is the nth number, n is prime. For example, 89 is prime, and the 11th Fibonacci number.

I don't quite get this one. The nth number of what sequence?
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Mike_Bson
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thicknavyrain wrote:
Mike_Bson wrote:
thicknavyrain wrote:
mdyrud wrote:The Fibonacci Sequence has an even number every third term?

Sonofa...

Also, if a Fibonacci number is the nth number, n is prime. For example, 89 is prime, and the 11th Fibonacci number.

I don't quite get this one. The nth number of what sequence?

Sorry, I meant to say:

Also, if a Fibonacci prime is the nth Fibonacci number, n is prime. For example, 89 is prime, and the 11th Fibonacci number.

skullturf
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mdyrud wrote:Even though the result is a unwieldy and probably not terribly useful, good job. One of the best ways to get better at math is exposing yourself to it, and it sounds like this is a really good way to give some great practice with rearranging equations. So don't worry about justifying yourself. Embrace your inner geek. After you've gotten bored of doing that, move on to proofs. The Fibonacci Sequence has an even number every third term? Not exactly a useful fact to know, but it's something to prove, and it gives you practice that will be useful in other, more practical problems.

I agree with this advice, and I apologize if my earlier remark seemed a little snotty. It's true that one of the best ways to get good at mathematics is simply to play around with stuff, and not to worry too much about whether what you come up with is already known or can be done more efficiently.

Lyon
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Dopefish wrote:I take it you just rearranged the quadratic formula?

Sort of. I knew that the quadratic formula is equal to x, so I took ax^2+bx+c=0, solved for x, and then substituted for x, and then I solved for each of the variables.

Nitrodon
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Mike_Bson wrote:Also, if a Fibonacci prime is the nth Fibonacci number, n is prime. For example, 89 is prime, and the 11th Fibonacci number.

There is exactly one number for which this statement does not hold.

Talith
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Lyon wrote:
Dopefish wrote:I take it you just rearranged the quadratic formula?

Sort of. I knew that the quadratic formula is equal to x, so I took ax^2+bx+c=0, solved for x, and then substituted for x, and then I solved for each of the variables.

You realise that solving for x means finding the quadratic formula yes? Just so we're clear, what definition are you using for 'solve for x'?

mr-mitch
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Nitrodon wrote:
Mike_Bson wrote:Also, if a Fibonacci prime is the nth Fibonacci number, n is prime. For example, 89 is prime, and the 11th Fibonacci number.

There is exactly one number for which this statement does not hold.

n != 4

Mike_Bson
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Nitrodon wrote:
Mike_Bson wrote:Also, if a Fibonacci prime is the nth Fibonacci number, n is prime. For example, 89 is prime, and the 11th Fibonacci number.

There is exactly one number for which this statement does not hold.

Ah yes, F(4) = 3.

Lyon
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Talith wrote:
Lyon wrote:
Dopefish wrote:I take it you just rearranged the quadratic formula?

Sort of. I knew that the quadratic formula is equal to x, so I took ax^2+bx+c=0, solved for x, and then substituted for x, and then I solved for each of the variables.

You realise that solving for x means finding the quadratic formula yes? Just so we're clear, what definition are you using for 'solve for x'?

I mean that I took the equation ax^2+bx+c=0, I solved it for x, and set it equal to the quadratic formula, so it looked like this.

Then I just started playing around with it

Dopefish
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Typically "solving for x" means isolating x, which is why many of us are/were perhaps a little confused when the quadratic formula didn't fall out when you solved for x. Perhaps as an excersise (if you haven't already) you should try to derive the quadratic formula from the the original ax^2+bx+c=0?

Also it may be worth pointing out that, your method of derivation has issues in the event that one of the roots happens to occur when x=0, not to mention theres a number of conditions you've imposed on the various constants that need not be true (for example that c=/=0, which isn't always true of a quadratic).

thicknavyrain
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Dopefish wrote:Typically "solving for x" means isolating x, which is why many of us are/were perhaps a little confused when the quadratic formula didn't fall out when you solved for x. Perhaps as an excersise (if you haven't already) you should try to derive the quadratic formula from the the original ax^2+bx+c=0?

Also it may be worth pointing out that, your method of derivation has issues in the event that one of the roots happens to occur when x=0, not to mention theres a number of conditions you've imposed on the various constants that need not be true (for example that c=/=0, which isn't always true of a quadratic).

just to give a little hint in the derivation, complete the square
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xepher
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Something interesting I found when I was messing with it as well. With some really simple manipulation, you get
2ax+b=sqrt(b^2-4ac)
And what's more is that 2ax+b is the derivative of ax^2+bx+c.

gorcee
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As an aside, some people may not know that there is an alternate form of the quadratic equation:

$x = \frac{2c}{-b \mp \sqrt{b^2-4ac}}$

In numerical applications, sometimes the standard quadratic equation will have cancellation in the computation of one of the roots caused by underflow in the calculation of [imath]b^2[/imath] or [imath]4ac[/imath], so using this formula will avoid that issue (of course, accuracy is almost always taken on credit; using the alternate formula will usually calculate the OTHER root correctly, but not the first. You must always pay the piper).

Ended
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xepher wrote:Something interesting I found when I was messing with it as well. With some really simple manipulation, you get
2ax+b=sqrt(b^2-4ac)
And what's more is that 2ax+b is the derivative of ax^2+bx+c.

This is an example of one of the properties of the discriminant. If a polynomial p(x) has a discriminant which is zero, then p has a multiple root, say at x0, and so the derivative p'(x0) is also zero. Quadratics only have two roots, so in this case the derivative at a root being zero is equivalent to the discriminant being zero.
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thicknavyrain
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In a mock university interview, I was asked some kind of question on the complex root of any quadratic in terms of the formula and there was something very interesting about it because one of the coefficients entirely disappeared but for the life of me I can't remember it now.
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Talith
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thicknavyrain wrote:In a mock university interview, I was asked some kind of question on the complex root of any quadratic in terms of the formula and there was something very interesting about it because one of the coefficients entirely disappeared but for the life of me I can't remember it now.

Do you mean that if a+bi is a root of a quadratic then so is a-bi? Put another way, if z is a complex number and solves the equation ax^2 + bx + c, then so does z*. It follows that the sum of the roots of any quadratic is always real (and the product).

thicknavyrain
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Talith wrote:
thicknavyrain wrote:In a mock university interview, I was asked some kind of question on the complex root of any quadratic in terms of the formula and there was something very interesting about it because one of the coefficients entirely disappeared but for the life of me I can't remember it now.

Do you mean that if a+bi is a root of a quadratic then so is a-bi? Put another way, if z is a complex number and solves the equation ax^2 + bx + c, then so does z*. It follows that the sum of the roots of any quadratic is always real (and the product).

No, I know about imaginary roots of a quadratic always coming in complex conjugate pairs, this was showing how the coefficients (such as the a, b and c in ax^2+bx+c=0) affect what the imaginary root will be and it turned out that either the c or the b didn't matter.
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Lothar
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thicknavyrain wrote:No, I know about imaginary roots of a quadratic always coming in complex conjugate pairs, this was showing how the coefficients (such as the a, b and c in ax^2+bx+c=0) affect what the imaginary root will be and it turned out that either the c or the b didn't matter.

It's pretty clear from the quadratic formula that if the roots of a quadratic with real coefficents are not real, then the real part of the roots is -b/2a, and hence is independent of c. The imaginary part is dependent on all three coefficients, however.
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1+1=3 for large values of 1.