Im in algebra II and we got this problem, the prof. said its harder than it looks so im just wondering

f(x)=x2+x3 and f(x)=0 find the numerical value of X

The numbers following x are exponents

Is the answer 0? I dont think it is though because he said its harder than it appears.

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If your saying f(x)=x^2+x^3=0 then yes, x does = 0. Then again x could be -1 where -1^2 = 1 and -1^3 = -1 which would give you 1+-1=0.

But ifyour saying f(x)=x^2+x^3 and find the value of x then the answer is undefined because X has no set value......

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I think it is zero. But I haven't encountered a problem like this in years, so I may be wrong.

I think we're learning about this tomorrow, so I'll have an answer then.


Edit- damn you Pr0az.

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wouldnt f(x)=0 define x?

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If f(x) does indeed = zero, then x3+x2 also =0. The only thing that can make this possible is if x=0. I think. It's been so long since I took a math class. I wonder what the fuck "f" represents? Wouldn't make a difference though.

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well f(x)=0 is defining it, but however is it

f(x)=x^2+x^3=0
f(x)=0

or x is defined as f(x)=0
so
x = 0

therefore what would
f(x)=x^2+x^3
so pluging in x would be
f(0)=0^2+0^3
f(0)=0

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Then again x could be -1 where -1^2 = 1 and -1^3 = -1 which would give you 1+-1=0. This would give you zero...but this is only if x is not already defined at zero.

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i'm drunk and haven't had any algebra for years..but i would say if f(x) = 0 and f(x) = x2 +x3 then obviously x2 + x3 = 0 and so x is 0 unless f and x are 0...then its a trick question...sorta.

f(0)=0^2+0^3
=0

f(-1)=(-1)^2 + (-1)^3
=1-1
=0

So your answer could be 0,-1

I remember this shit...It's easy.

I don't remember how to do it, and I can't even say I really looked at your question. lol.


Anyways, ask your teacher.

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Your contribution to this thread was legendary...

Thanks!


Yes, I'm pretty damn sure x=0 and you just plug it into the equation.


PAY ATTENTION IN CLASS NEXT TIME.

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The f(x) means its a function, which can also be denoted as y.

So basically you are finding what x can be to make the whole thing equal 0.

So yes, if you make x=0, the whole side is equal to 0 and therefore x=0 if f(x)=0.

Of course, -1 works as well, so it could possibly be a range of numbers, and thats why your teacher said it might be harder than it looks. There may be more but it's too late to be doing math.

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I seem to remember that an equation like this should strictly have three solutions, and I believe that since:

x^2(x+1)=0

The solutions are 0 twice, and -1

The "0 twice" is something I remember from maths a few years ago. Maybe John can shed some light on it?

From memory, the solution comprises one real and two (potentially) complex solutions. Solving cubics is a long process involving quite a few substitutions along the way (hence why this problem is in an algebra class) so I'll have a look at this when I get home from work or maybe this afternoon if there's a bit of a lull.

EDIT: Incidentally, everybody in this thread has "solved" the problem by inspection, rather than by proof.

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Do you mean using a proof? It just asks to find the value of X, not prove anything. If it's a function, it means it's a graph, which means X is a value.

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Proof meaning doing work, working the problem out, vice just looking at it and going oh thats 0 and or -1. i'm assuming thats what he meant.

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Oh.. Well, I guess that's because we figured he knew how to substitute in the numbers and check the math. He IS in Algebra

I just whipped out my good ol graphing calculator and using the table, it's 0 and -1 for sure. It extends up into infinity if you go higher or lower. Im just wondering if it's any more complicated, like a trick question or something

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Solving graphically, is still solving by inspection. The way to solve the problem is to derive a formula comprised entirely of the coefficients of the expression in question and then solving in a similar way as to the quadratic formula for quadratic equations.

I looked back over near 5 year old lecture notes last night and there's no way I'm going to reproduce the three and a half page derivation. The nub of the matter is that one needs to make a few carefully chosen substitutions to reduce the cubic to a quadratic, which can then be solved using the quadratic formula. But, there is a short cut that the good lecturer provided at the end and it goes as follows:

For equations of the form:

x^3 + ax^2 + bx + c = 0

Let:

(1)

Q = (3b - a^2)/9
R = (9ab - 27c - 2a^2)/54

and then:

(2)

D = Q^3 + R^2
S = cubedrt(R + sqrt(D))
T = cubedrt(R - sqrt(D))

The roots of the cubic equation are therefor given by:

x(1) = -a/3 + (S + T)
x(2) = -a/3 - (S + T)/2 + (i(sqrt(3))(S - T))/2
x(2) = -a/3 - (S + T)/2 - (i(sqrt(3))(S - T))/2

where i = sqrt(-1)

So for x^3 + x^2 = 0 we have in (1):

Q = -1/9 and;
R = -1/27

And in (2):

D = 0
S = -1/3
T = -1/3

So the solutions:

x(1) = -1/3 + (-2/3) = -1
x(2) = -1/3 + 1/3 + i(sqrt(3))*0 = 0
x(3) = -1/3 + 1/3 - i(sqrt(3))*0 = 0

So solution (1) is the real solution and solutions (2) and (3) are the complex solutions with the both the real and imaginary parts equal to zero.

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the answer is obviously 42

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Mansley totally pwnorz3d the formula.

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...which was my second guess.

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Yeah my face just melted

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Yeah i can see the aftermath in your avatar, i'm sorry for your loss of face.

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Oi! What a comedian!

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John Mansley quite makes it obvious why his tabs are so fucking good, with his math ability.

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I can't take all the credit - Cardano found the solution in the 16th century! I believe he also found the solution to the quartic. I think it was Gauss who found the solution to the quintic. Generally, if the index (highest power) of an equation is n, then there are n solutions.

Oh, I also forgot to mention that Danny got part way to the solution by factoring (a non-inspection technique) but it had to be assumed that the third solution was zero.

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Dude, you just brought my hangover on while I'm still drinking. Thanks alot.

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Fair enough.

Just something that is overlooked a lot... what is a polynomial to the 6th degree called (you verified 5th is quintic)... but what is 6? hexic (don't think so), sexic (likely), (sectic .. 7th?)... meh.

Damn maths is turning more and more into an English subject these days.

I think they would be hexic and heptic respectively, but mathematicians wouldn't use this terminology for post-quintic equations as they'd probably just say "a degree n equation" or "an equation of index n".

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See, I'm smart enough to be able to read all that, but understanding it is a completely different thing.

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Well, that's partly (along with a general unwillingness to spend a sizable chunk of my life reproducing it!) why I didn't post the derivation of the final formulae for x(1,2,3). It is quite hard to keep up with the substitutions and subsequently what terms the equation is expressed in (and, more importantly, getting back to the original terms) - it would just confuse most people and is probably over and above what the thread starter required for his class.

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the answer is 0 because u substitue 0 in for x. pretty much what u said with f(x)=o
F(0)=(o)2+(0)3=0

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Still solving by inspection!

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f(x)=0 does not define x to be zero, it defines some function, as expressed in terms of x, to be equal to zero. "f(x)=" is equivalent to using "y=", it's a simple matter of notation: Newton favoured f(x) whereas Leibnitz favoured setting the expression equal to another variable, usually y.

Just thought that needed clearing up.

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f(x)=0 walked into a pub and asked for a beer and a ham sandwich.
The barman replied....(wait for it you'll love this one)....."I'm sorry we don't cater for functions!"
Ha ha ha!







I really am so sorry for that!

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You seriously should be so

If
f(x) = x^2 + x^3
Can it not be easily written as
f(x) = (x^2) (1+x)
with solutions f(x) = 0 at either x^2 = 0 or (1+x) = 0 and with x to be either 0 or -1?

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Yes, factoring is a good way of solving equations but in normal circumstances it is very difficult to factor a cubic since they usually have complex solutions. And even if all three solutions are real, have any of you encountered algebraic long division? This is the only way to factor evil cubics (other than by inspection, of course) and is a nasty, nasty process (or at least I found it to be). It's much more efficient to use the formula.

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Factor Theorem still uses some guess and check, which by then you divide. The polynomials used for this ususally start out as factors, so can easily be factorized as an exercise.

Good luck trying to factorise

23x^4 + 3x^2 - 5x + 9482634 = 0

Hah.

Unless you're talking about more advanced algebraic / polynomial long division?



Asses (USS) stop reviving the thread without checking if the problem has been solved yet.

There's no guess work in algebraic long division just as there isn't in numeric long division. From memory anyway. But as I say, the formula is a lot less painless and guarantees the three solutions. Factoring and the subsequent long division is only more efficient if the formula is not known.

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Hmm, you're right about the division, but complete factorization (at my level) still requires some (methodical, educated) guess and check.

Hmm so you're a pure math freak? I talked to some math higher-ups about this and they said solving (by what you call) inspection is fine. The two zero roots and -1 should be adequate. Basically was questioning the necessity of such a massive derivation for this, when 'inspection' is fine.

Well, I studied mathematics to degree level and inspection isn't strictly the correct way to go about solving these equations, although it is a useful tool. It's like with quadratics; everybody uses the formula, right? [Or at least you should!]

The point your friends make about "the two zero roots" doesn't really hold as one has to assume that there are two roots at that point. Look at the graph of the equation in question: it intersects the x axis at -1 and becomes tangential at 0. It therefore, by inspection, appears to only have two solutions and so only knowledge that cubics have three solutions leads them to the assumption that 0 is a double root, which makes the assumption reasonable but does not prove it. [Even proving that a cubic has three solutions would only prove the second zero implicitly]

Mathematics is a very rigorous discipline and inspection is treated as a useful short cut that works some of the time, but it is never considered as the prime option. For example, if inspection was fine we wouldn't need mathematical induction as a method of proof.

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Math is so fun!!

I just learned how to do Synthetic Division in my Algebra II class.

I hate math! but its because my math-braincells are dead..btw i dont really think i ever had any haha!

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update of my algebra 2 class. FAILED. i get to take it again this quarter, with the same prof!

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lol Algebra is easy as hell

Not when your Prof is Greek and only knows about 10% of the English language, haha.

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It's not English, it's Maths.

Every, and I mean every, mathematical question can be posed without the need for English. It is done mostly via number, group and set theory with a healthy sprinkling of symbolism but it's often impractical to express problems in this form and so a combination of mathematical and English language is commonly employed.

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