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5:00 AM
oh missed a step
hm, you still have an x in there - can you get rid of that? (also, it might be worth keeping things in terms of "[stuff] = 0" for Reasons™)
Oh. Ohh
now, the equation should be much more manageable - how many solutions are there to t² - 3t + 1 = 0 ?
well it's not a perfect square, so two
wait... yup not a perfect square...
I think
that's not the only thing you have to check - it's not a perfect square, so either two or zero. how do you know it's two rather than zero?
5:02 AM
Because there's that +1
and 1=/=zero
sure, but "t² + 1 = 0" has no (real) solutions
what tells you how many solutions a quadratic has?
Oh I have to apply one of those stuff from algebra 2 I forgot
There's a simplified form of the quadratic formula
yep, it's one of those things!
Is b^2-4ac thing available at that level?
you don't need to actually calculate the full thing, but there's part of the quadratic formula that might help...
yes that
5:04 AM
Yes, I think it was that equation
that's what i was hinting at - the discriminant
do you remember what it tells you?
(also, that's an expression, not an equation! an equation is a statement "[stuff] = [other stuff]", and it can be true or false. an expression doesn't have an = sign -- it can't be true or false.)
It tells you whether the graph of a quadratic will have 0,1, or 2 roots
do you remember how it tells you that?
(Yeah, about half of mathematics is about getting the terminology correct)
Urm, I would be accidentally setting stuff up. It had to do with b and c though if I remember correctly
5:06 AM
@Bubbler (the other half is about getting it wrong)
not sure what you mean by "I would be accidentally setting stuff up"
but b^2-4ac is the discriminant -- it appears in the quadratic formula as "...±√(b^2-4ac)..."
well okay. If b^2-4ac>0, it has two roots. If b^2-4ac=0, it has exactly one root. If b^-4ac<0, it has no real roots
so, how many roots does "t² - 3t + 1 = 0" have?
5:08 AM
2, because -3^2-4(1)(1)>0
Okay, now let's get to the real question and see if I can deduce it in a similar fashion
It was -3^2, not 3
well, there's one more thing to note - you're not quite done
Nitpick: you should write (-3)^2, not -3^2
(also yes, that ^. we know what you meant but technically exponents come before the negative sign)
you know that there are two values of t that make it 0. but the question asked for values of x -- t was just something we made up to make things easier!
so, for any value of t, how many corresponding values of x are there?
5:10 AM
For every value of t, there is an "(x^3)" value of x?
No no
remember, t is just shorthand for x³
so if you know what t is, how many xs are possible that could give you that value of t?
(x)(x)(x). Urm
if you know "x³ is 8", how many possible values of x are there that would end up with that result?
5:12 AM
The question is not asking for that
No urgh
the cube root is how you would reverse it
i'm just asking you to count something here though
how many numbers are there whose cube is 8?
what numbers are they?
you've given me one number, but you said there were three of them
those are the same number
for comparison, if i asked "how many numbers are there whose square is 25", the answer would be "there are two: 5 and -5 both square to 25"
5:14 AM
i'm asking how many different numbers there are that would give you a result of 8 when you cube them
Just 1
That being 2
and as you mentioned before, you can do this more generally - no matter what value you have for t, you can always take the cube root and get a value for x
(another nitpick: more precisely, there are 3 of them, but two of them are not real)
5:15 AM
so every t value will give you exactly one x value. you know there are 2 t values that work, so...
(yes, i'm ignoring complex numbers at the moment)
There are exactly 2 x values
and there you go! now you're done
Hurray! Now for the real question
What, it isn't the end?
It was -3x^2, not -3x^3
I typoed
5:16 AM
that extra thing didn't actually matter here, but it might in some other scenario where you couldn't get exactly one every time
x^6-3x^2+1=0. Can we apply the same principal. Lemme see
This time, it's t^3-t+1
no quadratics. rats. however, that looks like a special factor thingy
unfortunately, i... don't believe it is
oh darn
craaap more time i need to sleep soon
Whatever, this is genuinely the last question
not sure what specifically they want you to do here - you can do the same technique but it's a bit more painful
(Spoiler using calculus: t^3-t+1=0 has only one real root, and it is negative, so the original equation has zero real roots.)
5:19 AM
There's Descarte Rule of signs still
(it's -3t, not -t)
maybe they expect some form of IVT argument?
yeah that's my guess, but without a graph that seems weird
5:20 AM
3t makes three roots, and you can indeed use Descartes for that
because descartes only gives you the parity of the number of roots of each sign
I interpreted the question as "don't use graph for justification", not "don't use graphs at all"
Idk what they intended
hm that's possible
Use Descartes at t = -2, 0, 1, 2.
That gives - + - + which gives the maximum of 3 roots
Not sure what that means?
5:22 AM
huh? No
How do you use Descartes at a point
also confused
it's x^6-3x^2+1
well, you've "simplified" it to t³-3t+1
Same principle
That means that there are a maximum of 2 roots for positive
5:23 AM
right, so the question is whether there are 2 or 0
And negative
oh wait this argument can work actually
hm no never mind not without a bit more
sorry, go on?
Oh, it's not Descartes
It can be 2,1, or zero
Wait not one
2 or zero
Forgot about conjugates for a second
5:25 AM
yeah, i'm guessing that the intended argument is just to actually show there are positive roots by IVT or something? but ew
What's IVT?
intermediate value theorem
not sure what theorems you have 'access' to
Oh wait we learned this theorem but I wasn't paying attention
Yeah, I think I meant IVT
5:26 AM
Almost ten years since calculus :/
f(a)-f(b)/f(a-b)=c or something??
@Bubbler ah ok yes
@PrinceNorthLæraðr that's MVT
IVT is much simpler
Mean Value Theorem
@Deusovi Oh it's the one where C must be somewhere between a and b
5:27 AM
The gist is that if f(a)<0 and f(b)>0, there is a root between a and b
So... I plug in two numbers?
Basically if f changes sign from a to b, there's a root between a and b
Two numbers to prove one root is there
So if you can find a,b so that f(a) and f(b) have different signs. there's definitely a root between them
5:29 AM
Four numbers to prove three roots are there
what numbers do I pick though?
that's the tricky part
like, you can just graph and look
but i'm not sure how they intend you to find them
how about no graph?
Just keep trying the "simplest" numbers you can think of?
Usually the rule of thumb is to try small positive/negative integers
5:30 AM
I'm guessing they don't want you to graph
so 1 and -1
like -2, -1, 0, 1, 2
yeah just trying simple numbers is a good approach when all else fails
you're looking for positive roots though so you want to keep your numbers positive (or 0)
"there's a root between -1 and 1" doesn't help you find positive roots, because it might be -1/2 or something
But if you don't identify the intervals of all three roots, you don't get conclusive results for the original problem
not necessarily bubbler
wdym sign changw?
5:32 AM
we know there's one negative root, so the question is whether there are 0 or 2 positive roots
Ok then
Then the numbers are simpler: try just 0, 1, 2
@PrinceNorthLæraðr Like if at 0, the value is say -20, at 1 the value is 30, so then you know there's a root between 0 and 1
@Deusovi ah, not necessarily
x^6-3x^2+1 has different values in Descarte Rule of Signs than t^3-3t^1+1
5:34 AM
right now we're just looking at t³-3t+1
@PrinceNorthLæraðr You don't need to apply Descartes to the original equation
the whole point of this is that you're finding where t³-3t+1=0, and then trying to see if the results you get from that give you anything for the original
@Bubbler Well, I kind of feel like that might've been the point, just really badly worded
I just graphed x^6-3x^2+1=0, and it has four roots, 2 negative, 2 positive, which ta-da fits the descarte rules with 2 complex
descartes can't help you very much there - you can't easily prove the number of roots of the original, even using both descartes and IVT
5:36 AM
it fits, but other numbers fit too
descartes only tells you the parity of the number of roots so it's not super helpful
There must be an easier solution- the graph of x^6-3x^2+1 is oddly very symmetrical
It is symmetrical because it is also a polynomial in t=x^2
i.e. all terms have even powers
It looks almost identical to an x^4 graph
right, since all the powers of x are even, it's going to be an even polynomial so it'll be symmetric
Huh didn't know that
Wait that means that descarte could work, no?
Wait no
5:39 AM
you could use descartes, symmetry of even polynomials, and IVT all together
but that sounds like a pain
It's possible the graph could've just looked like an x^2 graph right
with different coefficients, yeah
honestly i'm still not sure what they exactly want here
@Deusovi That sounds way too much for Pre-Calc
"How many real solutions"...
it is just things you know (or at least i assume things that have been taught at one point)
but it's a lot to do all at once
I don't remember IVT being in this unit. It was last unit. And I don't think we learned about even symmetry
ugh, i'm so tired
ivt was prob on this unit actually
5:43 AM
even then you'd still need to get lucky enough to pick points that give you a sign change for ivt
well, (0,1) worked
but what if it didnt?
right, that's why we were hesitant too
they happen to be small but if they weren't you'd just be out of luck
If a question is intended to use IVT, I bet it is always designed to work with simple numbers
right, but the question is whether IVT was intended
5:46 AM
Honestly I don't think you can prove there are exactly 4 roots without IVT or calculus
welp asking my teacher tomorrow
oh crap i missed a question
Forget it
It's 11 pm here and I need to get up at 6 am
I'm going to sleep
Thanks everyone for the help!
good night!
Good night!
Well, it's almost done (assuming IVT is allowed) - we've got t^3-3t+1=0 has two or zero positive roots, and we've got one positive root is in (0,1), so it has two positive roots. Then t=x^2, which has two roots per positive t, so the original equation has four roots.
At this point, we can't distinguish mathematics from puzzling.
It feels too hit-or-miss for a homework problem ig
6 mins ago, by Bubbler
If a question is intended to use IVT, I bet it is always designed to work with simple numbers
5:53 AM
Guess so
If you plug in some simple numbers to use IVT and it doesn't work, you can quickly abandon IVT on that equation.
But that's kinda quirky problem solving technique after all
(which works only until you get to basic calculus, where you can identify local minima/maxima and etc directly)
The calculus way: Identify t=x^2 substitution. Identify that local minima/maxima are at t=-1 and t=1, and the graph has rotational symmetry around t=0. Plug in the three values to guess the shape of the graph and derive that two roots are positive and one is negative. Plug in to t=x^2 to conclude that there are four real roots.
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