freaking_out
Saddam's new life
1. prove that the polynomial x^3+ax^2+bx+c has a double root if and only if (9c-ab)^2=4(a^2-3b)(b^2-3ac)
thanks :worried:
thanks :worried:
Q. a polynomial question (1 Viewer)
- Thread starter freaking_out
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turtle_2468
Member
OK. Part of this will be an outline, because you know this is a hard question, so be prepared to do the bashing 
don't worry, the bits I leave out are just expanding algebra.
For the only if bit: assume roots x1, x1, x2. Then you get the three equations such as (x1^2)*x2=-c. You could sub them in at this point and get something really horrible, but I suggest expanding and cancelling. This means that you can divide all the terms by c. To make it even nicer, divide by b and note that (from the sums, products etc of roots, and cancelling) =(2/x1)+(1/x2).
The if bit: Much shorter to write but harder to understand. Consider any "general" cubic (ie ones that don't have a triple root... those you can prove quite easily anyway in another case). Then let the x-coords of its turning points be at k and l. Suppose that the algebraic condition working (*) leads to no double root at x=m.
THE TRICKY BIT: (really!) Consider the cubic x^3 etc with a and b FIXED but c VARIABLE. (let the original value of c given be c'... this will come in handy in the next bit)That means that you can move the cubic up and down on the number plane.
Then you know that the given algebra equation (9c-ab etc...) is satisfied by c=c' (from *). But furthermore, the cubic has TWO turning points with y-coordinates not at zero! (as there are no dbl roots) Let these turning points have y-coordinates p,q, neither of them zero.. Then shifting the graph down by p and q respectively (ie setting c=c'-p and c=c'-q) (if you have trouble underestanding this, remember that c is the "constant" value in the cubic graph), we see from the "only if" part that c=c'-p and c=c'-q also result in the algebraic equation being satisfied. So we have 3 distinct values of c satisfying the equation.
BUT the equation considered in the variable c is only a quadratic and so cannot have 3 solns! So we have a contradiction and hence end the proof.
Happy to explain any of that ppl didn't get...
don't worry, the bits I leave out are just expanding algebra. For the only if bit: assume roots x1, x1, x2. Then you get the three equations such as (x1^2)*x2=-c. You could sub them in at this point and get something really horrible, but I suggest expanding and cancelling. This means that you can divide all the terms by c. To make it even nicer, divide by b and note that (from the sums, products etc of roots, and cancelling) =(2/x1)+(1/x2).
The if bit: Much shorter to write but harder to understand. Consider any "general" cubic (ie ones that don't have a triple root... those you can prove quite easily anyway in another case). Then let the x-coords of its turning points be at k and l. Suppose that the algebraic condition working (*) leads to no double root at x=m.
THE TRICKY BIT: (really!) Consider the cubic x^3 etc with a and b FIXED but c VARIABLE. (let the original value of c given be c'... this will come in handy in the next bit)That means that you can move the cubic up and down on the number plane.
Then you know that the given algebra equation (9c-ab etc...) is satisfied by c=c' (from *). But furthermore, the cubic has TWO turning points with y-coordinates not at zero! (as there are no dbl roots) Let these turning points have y-coordinates p,q, neither of them zero.. Then shifting the graph down by p and q respectively (ie setting c=c'-p and c=c'-q) (if you have trouble underestanding this, remember that c is the "constant" value in the cubic graph), we see from the "only if" part that c=c'-p and c=c'-q also result in the algebraic equation being satisfied. So we have 3 distinct values of c satisfying the equation.
BUT the equation considered in the variable c is only a quadratic and so cannot have 3 solns! So we have a contradiction and hence end the proof.
Happy to explain any of that ppl didn't get...
N
ND
Guest
Here's a similar question with solutions:
http://www.boredofstudies.org/community/showthread.php?s=&threadid=5129&highlight=*polynomial*
http://www.boredofstudies.org/community/showthread.php?s=&threadid=5129&highlight=*polynomial*
N
ND
Guest
Turtle: there are a few parts that i don't understand.
If you would explain any of that, i would appreciate it. Thanks.
What exactly are you expanding (then dividing by c and b)?
What is the significance of x=m?
I don't understand how this shows (9c-ab)^2=4(a^2-3b)(b^2-3ac).
If you would explain any of that, i would appreciate it. Thanks.
turtle_2468
Member
What exactly are you expanding (then dividing by c and b)?
Let's call the algebraic condition (9c-ab etc) be called (**) for easy notation.
Basically, proving the "only if" bit (that if you have a dbl root then algebraic cond holds) is a big bash. I was going to expand (**), and then try to bash that out... sorry, I'm just a little lazy I was just saying though that after you exapdn it and simplify the terms (divide by 3 while you're at it) you can then divide 3a^2b^2+54abc-12a^3c.... (that equation you get when you expand) by bc, which should make the algebra slightly nicer. You could make it even nicer by substituting in (2/x1+1/x2)=-b/c, where x1 is the dbl root, x2 is the single root, and you get that identity by dividing {(x1^2+2x1x2)=b} by {x1^2*x2=c}.
What is the significance of x=m?
Nothing... whoops! I'm doing a proof by contradiction, so that bit just means: "assume there isn't a double root"
I don't understand how this shows
(**).
OK. This is the if part, so we have to show that if the algebraic condition holds then there is a dbl root... (the "if" part is very bashy...
Short note about the "philosophy" of the proof: We first assume that there isn't a double root there. Then we show that there are 3 values of c which satisfy (**), which is a contradiction as the eq is a quadratic in c which can have at most 2 solns.
OK, that didn't make too much sense. Intuitive/geometric proof: (READ THIS IF YOU DON'T GET MY METHOD) changing c (remember sketching graphs) is equivalent to shifting the whole graph up or down. When you change c to say c=q_1 (q_1 a constant) so that a turning point is on the x-axis, ie a double root exists, then c=q_1 satisfies the algebraic condition by the "only if" part. As cubics generally have two double roots, there is another turning point you can put on the x-axis with different y-coordinate, with c=q_2, so that (**) is again satisfied.
Now assume that there is a third value, c=q_3 (what they GIVE YOU as c, ie the original value), such that the (**) holds but there is NO DOUBLE ROOT. Then as there is no double root, q_3 is not equal to either q_1 or q_2. So there are 3 values of c, c=q_1, c=q_2, and c=q_3 that satisfy (**), which is a contradiction as (**) is only a quadratic in c.
Example to follow if you still didn't get that... I know it's kinda abstract, so sorry if I didn't explain it well
Let's call the algebraic condition (9c-ab etc) be called (**) for easy notation.
Basically, proving the "only if" bit (that if you have a dbl root then algebraic cond holds) is a big bash. I was going to expand (**), and then try to bash that out... sorry, I'm just a little lazy
I was just saying though that after you exapdn it and simplify the terms (divide by 3 while you're at it) you can then divide 3a^2b^2+54abc-12a^3c.... (that equation you get when you expand) by bc, which should make the algebra slightly nicer. You could make it even nicer by substituting in (2/x1+1/x2)=-b/c, where x1 is the dbl root, x2 is the single root, and you get that identity by dividing {(x1^2+2x1x2)=b} by {x1^2*x2=c}. What is the significance of x=m?
Nothing... whoops! I'm doing a proof by contradiction, so that bit just means: "assume there isn't a double root"
I don't understand how this shows
(**).
OK. This is the if part, so we have to show that if the algebraic condition holds then there is a dbl root... (the "if" part is very bashy...
Short note about the "philosophy" of the proof: We first assume that there isn't a double root there. Then we show that there are 3 values of c which satisfy (**), which is a contradiction as the eq is a quadratic in c which can have at most 2 solns.
OK, that didn't make too much sense. Intuitive/geometric proof: (READ THIS IF YOU DON'T GET MY METHOD) changing c (remember sketching graphs) is equivalent to shifting the whole graph up or down. When you change c to say c=q_1 (q_1 a constant) so that a turning point is on the x-axis, ie a double root exists, then c=q_1 satisfies the algebraic condition by the "only if" part. As cubics generally have two double roots, there is another turning point you can put on the x-axis with different y-coordinate, with c=q_2, so that (**) is again satisfied.
Now assume that there is a third value, c=q_3 (what they GIVE YOU as c, ie the original value), such that the (**) holds but there is NO DOUBLE ROOT. Then as there is no double root, q_3 is not equal to either q_1 or q_2. So there are 3 values of c, c=q_1, c=q_2, and c=q_3 that satisfy (**), which is a contradiction as (**) is only a quadratic in c.
Example to follow if you still didn't get that... I know it's kinda abstract, so sorry if I didn't explain it well
turtle_2468
Member
EXAMPLE: {THIS WILL COME IN USEFUL LATER}Consider P(x)=x^2*(x-2). Then expanding, this is same as P(x)=x^3-2x^2. THe turning points are (0,0) and (2,-4).
Onto the question proper. Let a, b be fixed, and c variable. suppose we have P(x)=x^3-2x^2+c.
Then (**) is equivalent to:
81c^2=4(4)(6c)
ie 81c^2=96c
ie 27c^2-32c=0... (@@)
(the algebra isn't important here... trust me, the approach works for more "complex" ones as well but it's harder to see the logic behind it then)
Suppose we have c=k which satisfies (**) (see previous post) but has no double root.
Then k is not 0 or 4. (when c=0, there are turning points at 0 and -4, so setting k at 0 or 4 "shifts" those turning pts onto the x-axis)
But then (@@), a quadratic in the variable c, has three solns: c=0, c=4 (these two from the "only if" part)
and c=k (from our assumption) CONTRADICTION as (@@) is a quadratic
So there can't be another k. So when you choose c such that (**) holds, it must be 0 or 4, so there must be a dbl root.
Onto the question proper. Let a, b be fixed, and c variable. suppose we have P(x)=x^3-2x^2+c.
Then (**) is equivalent to:
81c^2=4(4)(6c)
ie 81c^2=96c
ie 27c^2-32c=0... (@@)
(the algebra isn't important here... trust me, the approach works for more "complex" ones as well but it's harder to see the logic behind it then)
Suppose we have c=k which satisfies (**) (see previous post) but has no double root.
Then k is not 0 or 4. (when c=0, there are turning points at 0 and -4, so setting k at 0 or 4 "shifts" those turning pts onto the x-axis)
But then (@@), a quadratic in the variable c, has three solns: c=0, c=4 (these two from the "only if" part)
and c=k (from our assumption) CONTRADICTION as (@@) is a quadratic
So there can't be another k. So when you choose c such that (**) holds, it must be 0 or 4, so there must be a dbl root.
N
ND
Guest
Ah i understand now. Thanks heaps.
PS. You topped the state right?
PS. You topped the state right?
turtle_2468
Member
umm.. yeah.
freaking_out
Saddam's new life
wow, your good, how'd you get so good in 4u maths anyway?
how'd you get so good in 4u maths anyway?
turtle_2468
Member
umm... well I think there are 2 things that helped in the exam proper... bashing and being careful.
Bashing meaning that if you don't get, for example, a nice soln to an algebraic formula, just massively expand it
Being careful: Well something you try to do anyway. But if you have a little bit of spare time, just look over the q's. Also helps to do a "reality check" right after each question to make sure, for example, you haven't integrated something which is obviously positive and come up with a negative answer.
Bashing meaning that if you don't get, for example, a nice soln to an algebraic formula, just massively expand it
Being careful: Well something you try to do anyway. But if you have a little bit of spare time, just look over the q's. Also helps to do a "reality check" right after each question to make sure, for example, you haven't integrated something which is obviously positive and come up with a negative answer.
N
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Guest
Being good enough to go to the IMO helps a little too.Originally posted by turtle_2468
umm... well I think there are 2 things that helped in the exam proper... bashing and being careful.
Bashing meaning that if you don't get, for example, a nice soln to an algebraic formula, just massively expand it
Being careful: Well something you try to do anyway. But if you have a little bit of spare time, just look over the q's. Also helps to do a "reality check" right after each question to make sure, for example, you haven't integrated something which is obviously positive and come up with a negative answer.
turtle_2468
Member
indeed. lol, it was a lot of work, but fun..
freaking_out
Saddam's new life
hey, what does IMO stand for again????
N
ND
Guest
http://imo.wolfram.com/Originally posted by freaking_out
hey, what does IMO stand for again????