Note that when you have x⁴ + 4x² + x + 1 = 0 you can rewrite this as x⁴ = −(4x² + x + 1) x⁴ = −((2x + ¹⁄₄)² − ¹⁄₁₆ + 1) x⁴ = −((2x + ¹⁄₄)² + ¹⁵⁄₁₆) The left hand side is nonnegative for any real x, and the right hand side is negative for any real x, so there can be no real solutions for this quartic equation.
Your voice and speed make things easier. The way you explained and those steps you arranged to digest the problem, that is what ideal teacher did. I am medical personnel yet I can easily follow what you are doing, well with some self research and basic calculus, which I got C, in university time. Anyway I think many teachers should learn how to explain just like you do.😊
Amazing video bro, in the beginning I was thinking about solving the real quartic head on but then realized those roots are going to be irrational since the rational root theorem failed and then said there must be a easier way and then remembered Vietas formula. If the roots were rational then squaring them and adding them up would have been too easy. Also I want to point out that the intermediate value theorem only applies to continuous functions for those who don’t know.
Any polynomial is divisible by x-r. Write p(x)=a(n)*x^n+a(n-1)*x^(n-1)+...+a2*x²+a1*x+a0=an(x-r1)(x-r2)...(x-rn). Distribute the parentheses on the right side and Bob's your mother's brother, it works for any power.
I love problems like this, where (as in this example) you don’t find the actual roots r1, r2, r3, r4 but (without actually knowing those roots) you still find the value of some _function_ of those roots.
When you got (x^4+4x^2+x+1)(x^4-4x^2-x+1), you could find the discriminant of each polynomial, and find that x^4+4x^2+x+1 has two pairs of non-real complex conjugate roots, and x^4-4x^2-x+1 has four real and distinct roots. From here you could either solve for x^4-4x^2-x+1, or do what you did
as we know this problem can be solved like this we know (A+B) ^2= A^2+B^2+2AB (A+B+C) ^2= A^2+B^2+C^2+2(sum taken two at a time) (A+B+C+D...............+) ^2 - 2(sum take two at a time) = A^2+B^2+C^2+......... (i) from quadratic -(coefficient of x^7/coefficient of x^8) = sum taken at a time (coefficient of x^6/coefficient of x^8) = sum taken two at a time plugging values into equation (i) we get = 0
You can take the derivative of the quartic to get a cubic and then find it's zeros (by manipulation or by the formula for roots of a cubic). This will essentially give you where the curve of the quartic turns, which will tell you how many real roots it has. Then, as someone pointed out, use Newton's sums to get sum of any power of roots. I didn't know about this.
Mr. Prime Newtons : As mentioned by many below your style of explanation and delivery are really nice - very measured and clear. Well done ! This is an interesting problem, and you do justice to it. However, the part that leaves me somewhat unsatisfied about this problem as a contest question is the necessary first step of factoring this 8th degree polynomial into two quartics. One could spend a LOT of time trying to factor it thusly until one hits upon the elegant method of grouping and completing the squares that you show. Easy to see in retrospect, but quite difficult to accomplish going forward. ( Now, if the problem had a hint like : "Start by factoring into two quartics" that would give the would-be solvers a chance ). BTW, in the Harvard-MIT Math Tourney are these team questions or questions to be solved by individuals ? Also, how much time are they allotted ? Thank you !
Good job, man! I had a hard time factoring the original polynomial into two quartics and did not succeed. Of course I tried rational roots, but that gave nothing. Then I tried my own generic "Ansatz" with 6 unknowns and got lost trying. But you saw some things in the coefficients that really did it.
Love your videos! I was curious, as I couldn't really find anything particularly useful googling, is it at all possible to differentiate a tetrated function, where the base is the constant and the variable is the superexponent?
*New question…* Suppose that we are given a cubic (degree 3) polynomial, and, when graphed, the polynomial exhibits the shape of a capital letter N. Now suppose that the bottom-left of the N crosses the x-axis a little to the right of the origin, and the very bottom of the bottom-right of the N BARELY TOUCHES the x-axis. How many roots would this type of cubic polynomial actually possess?
When a polynomial curve "just barely touches the x-axis", i.e. when it is TANGENT to the x-axis, then that implies a double real root at that location. So, to answer your "N-shaped" curve question : That cubic polynomial would have 3 real roots ( all positive by the way you described it ) and the two roots farthest to the right would be exactly equal to each other ( hence the term "double root" ).
@@GWaters-xr1fv That is the answer that I was expecting that academia would endorse, however, such an answer should really undergo further review. Out of the 3 possible roots, if two of those roots are guaranteed to always be equal to each other (whenever the given assumptions happen to be true), then, in truth, such a degree-3 polynomial only has 2 roots (both of which are Real numbers.) When a degree-3 polynomial only has 2 roots, and both of the roots are Real numbers, I would prefer if academia would refer to such an instance as a depressed or degenerative case. In any case, thank you for your informed reply.
@@GWaters-xr1fv Didn't it say barely touches? How do you know if that's a double root? What if it just goes above the x-axis? In that case wouldn't it be just one positive real root?
@@abhirupkundu2778 When someone says "barely touches" they usually mean that it does touch, but barely ( i.e. does not cross ). If it comes close but doesn't actually touch then they wouldn't say "barely touches" - they might say "comes close to touching" i.e. does NOT touch. However you are correct in saying that if the cubic does not touch there, then there would only be one real root.
Is there any info to be gotten if you take the derivative of the quartic that provides the real roots and set it to zero to find the max and min points?
Sure, but the derivative P'(x) = 4x³ − 8x − 1 is a cubic which has three real roots and which can therefore only be solved trigonometrically (or numerically). Not really worth it. But, I agree that the explanation in the video about the number of real zeros of the polynomial P(x) = x⁴ − 4x² − x + 1 may not have been entirely clear if you haven't seen something like this before. The purpose of rewriting the polynomial as P(x) = x⁴(1 − 4/x² − 1/x³ + 1/x⁴) is to note that | −4/x² − 1/x³ + 1/x⁴ | < 1 for any sufficiently large |x| so −1 < −4/x² − 1/x³ + 1/x⁴ < 1 and therefore 0 < 1 − 4/x² − 1/x³ + 1/x⁴ < 2 for any sufficiently large |x| so P(x) will be _positive_ for any sufficiently large |x|. And since P(−1) = −1, P(0) = 1, P(1) = −3 this means there will be at least one zero on each of the four intervals (−∞, −1), (−1, 0), (0, 1), (1, ∞) And since there can be no more than four zeros, the conclusion follows that P(x) has _exactly one zero_ on each of these four intervals.
Good thought, but that wouldn't really address the question of the roots. For example, for a quartic polynomial ( with leading coefficient positive ) to have 4 real roots ( as does the quartic that Mr. Prime Newtons considers here ) it is certainly NECESSARY that it also has 1 maximum and 2 minimums. But, conversely, that is NOT a sufficient condition for having 4 real roots. i.e. having 1 max and 2 mins does not guarantee 4 real roots, and in fact it does not guarantee any real roots ! To see this, imagine that the final numerical term in this quartic was the number 10 instead of 1. That would simply shift the graph UP by nine units. It would still have 1 max and 2 mins, but would not have any real roots.
No. The function you give can be written as f(x) = (x − 2)(x − 3) so we have f(2) = 0 and f(3) = 0. The graph of your function crosses the x-axis at x = 2 and at x = 3. The value of your function is negative for any real x between 2 and 3. You can graph your function online at the desmos website.
A root of an equation is the root of a sulotion to the equation, hm? I'm a little puzzled, cause I might not have had this in math. What is the point of taking a root of an equation? Do you also take log or tan of an equation?
The graphical arguments [and positive infinity and negative infinity] was awesome, but it involved a mature discussion, maybe not suited to younger members of your channel.
@@Arkapravo sure, if you're 6 it's not normal to do that, but if you're watching videos of Harvard-MIT tournaments you should be expecting *something* clever to occur.
![Find (x+y+z) [Harvard-MIT] Guts contest](http://i.ytimg.com/vi/Pe0s5N_jkLA/mqdefault.jpg)
Note that when you have
x⁴ + 4x² + x + 1 = 0
you can rewrite this as
x⁴ = −(4x² + x + 1)
x⁴ = −((2x + ¹⁄₄)² − ¹⁄₁₆ + 1)
x⁴ = −((2x + ¹⁄₄)² + ¹⁵⁄₁₆)
The left hand side is nonnegative for any real x, and the right hand side is negative for any real x, so there can be no real solutions for this quartic equation.
That's smart 👌
Much more direct than the case-by-case analysis. Cool!
Once you have established the quartic polynomial with 4 real roots, you can also use newton's sum to obtain the sum of the squares of the roots.
Thank you for sharing. I never knew it existed.
JEE student?
I also studied it during JEE preparation
Your voice and speed make things easier. The way you explained and those steps you arranged to digest the problem, that is what ideal teacher did. I am medical personnel yet I can easily follow what you are doing, well with some self research and basic calculus, which I got C, in university time.
Anyway I think many teachers should learn how to explain just like you do.😊
Amazing video bro, in the beginning I was thinking about solving the real quartic head on but then realized those roots are going to be irrational since the rational root theorem failed and then said there must be a easier way and then remembered Vietas formula. If the roots were rational then squaring them and adding them up would have been too easy. Also I want to point out that the intermediate value theorem only applies to continuous functions for those who don’t know.
This is mind blowing
Thanks!
Thank you!
You made this problem look so easy! I would have to do many examples like this one to develop the skills needed.
the vieta's formula is fire . do a proof of it pleaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaase.
Any polynomial is divisible by x-r. Write p(x)=a(n)*x^n+a(n-1)*x^(n-1)+...+a2*x²+a1*x+a0=an(x-r1)(x-r2)...(x-rn). Distribute the parentheses on the right side and Bob's your mother's brother, it works for any power.
At 11:00 you can rewrite x^4+4x^2+x+1 as x^4 + (x+2)^2/4 + 15x^2/4 which is obviously always positive
is that he should do in order the case1 and 2
After factoring the initial
polynomial, you can apply
Decade's rule of signs to
see there are two positive
and two negative real roots.
All videos are outstanding
I would have glossed over the word “real”, written 14² - 2, and wondered why I had so much time left.
The answer would be 0, not 14^2 - 2, even if glossing over the word 'real,' via Vieta's formulae.
I love problems like this, where (as in this example) you don’t find the actual roots r1, r2, r3, r4 but (without actually knowing those roots) you still find the value of some _function_ of those roots.
When you got (x^4+4x^2+x+1)(x^4-4x^2-x+1), you could find the discriminant of each polynomial, and find that x^4+4x^2+x+1 has two pairs of non-real complex conjugate roots, and x^4-4x^2-x+1 has four real and distinct roots. From here you could either solve for x^4-4x^2-x+1, or do what you did
as we know this problem can be solved like this
we know (A+B) ^2= A^2+B^2+2AB
(A+B+C) ^2= A^2+B^2+C^2+2(sum taken two at a time)
(A+B+C+D...............+) ^2 - 2(sum take two at a time) = A^2+B^2+C^2+......... (i)
from quadratic -(coefficient of x^7/coefficient of x^8) = sum taken at a time
(coefficient of x^6/coefficient of x^8) = sum taken two at a time
plugging values into equation (i) we get = 0
Would love to see a video to solve the equation
You can take the derivative of the quartic to get a cubic and then find it's zeros (by manipulation or by the formula for roots of a cubic). This will essentially give you where the curve of the quartic turns, which will tell you how many real roots it has.
Then, as someone pointed out, use Newton's sums to get sum of any power of roots. I didn't know about this.
Very good! I would have never gotten past the first aha! moment where you broke up the 14 term as 16 and 2. Brilliant!
x^4 +4x^2 +x+1=x^4 +3.75x^2 +(x^2 /4 + 2 x/2 +1)=x^4 +3.75x^2 +(x/2+1)^2
Sir kindly upload videos on advance analysis
Hello there, from Russian Olympiad Community! Love you videos very much❤
Amazing teacher
Awesome problem explained in awesome way on how to solve it.
Mr. Prime Newtons : As mentioned by many below your style of explanation and delivery are really nice - very measured and clear. Well done ! This is an interesting problem, and you do justice to it. However, the part that leaves me somewhat unsatisfied about this problem as a contest question is the necessary first step of factoring this 8th degree polynomial into two quartics. One could spend a LOT of time trying to factor it thusly until one hits upon the elegant method of grouping and completing the squares that you show. Easy to see in retrospect, but quite difficult to accomplish going forward. ( Now, if the problem had a hint like : "Start by factoring into two quartics" that would give the would-be solvers a chance ). BTW, in the Harvard-MIT Math Tourney are these team questions or questions to be solved by individuals ? Also, how much time are they allotted ? Thank you !
Good job, man! I had a hard time factoring the original polynomial into two quartics and did not succeed. Of course I tried rational roots, but that gave nothing. Then I tried my own generic "Ansatz" with 6 unknowns and got lost trying. But you saw some things in the coefficients that really did it.
Love your videos! I was curious, as I couldn't really find anything particularly useful googling, is it at all possible to differentiate a tetrated function, where the base is the constant and the variable is the superexponent?
we finally got coloured chalk on this channel
Thank you Sir
Just wonder for clear explanation how to find solution, trying follow the lessons in past and go for future. Thanks a lot for your time, Sir.
Can you use big-O notation in a proof? O(x^2) > O(x) for x>1 and x
Can't we plugin directly?
We don't have what to plug in
Very nice and intelligent question
0? Oh right that was the sum of roots to 1st power. Then you showed sum of products of roots to derive the answer.
*New question…*
Suppose that we are given a cubic (degree 3) polynomial, and, when graphed, the polynomial exhibits the shape of a capital letter N. Now suppose that the bottom-left of the N crosses the x-axis a little to the right of the origin, and the very bottom of the bottom-right of the N BARELY TOUCHES the x-axis. How many roots would this type of cubic polynomial actually possess?
When a polynomial curve "just barely touches the x-axis", i.e. when it is TANGENT to the x-axis, then that implies a double real root at that location. So, to answer your "N-shaped" curve question : That cubic polynomial would have 3 real roots ( all positive by the way you described it ) and the two roots farthest to the right would be exactly equal to each other ( hence the term "double root" ).
To add : Tangency always requires a double-root, even when it is, say, a circle that is tangent to a parabola.
@@GWaters-xr1fv That is the answer that I was expecting that academia would endorse, however, such an answer should really undergo further review. Out of the 3 possible roots, if two of those roots are guaranteed to always be equal to each other (whenever the given assumptions happen to be true), then, in truth, such a degree-3 polynomial only has 2 roots (both of which are Real numbers.) When a degree-3 polynomial only has 2 roots, and both of the roots are Real numbers, I would prefer if academia would refer to such an instance as a depressed or degenerative case. In any case, thank you for your informed reply.
@@GWaters-xr1fv Didn't it say barely touches? How do you know if that's a double root? What if it just goes above the x-axis? In that case wouldn't it be just one positive real root?
@@abhirupkundu2778 When someone says "barely touches" they usually mean that it does touch, but barely ( i.e. does not cross ). If it comes close but doesn't actually touch then they wouldn't say "barely touches" - they might say "comes close to touching" i.e. does NOT touch. However you are correct in saying that if the cubic does not touch there, then there would only be one real root.
Тут главное - знать теорему Виета для полиномов степени больше 2.
great video!
0?
Is there any info to be gotten if you take the derivative of the quartic that provides the real roots and set it to zero to find the max and min points?
Sure, but the derivative P'(x) = 4x³ − 8x − 1 is a cubic which has three real roots and which can therefore only be solved trigonometrically (or numerically). Not really worth it.
But, I agree that the explanation in the video about the number of real zeros of the polynomial
P(x) = x⁴ − 4x² − x + 1
may not have been entirely clear if you haven't seen something like this before. The purpose of rewriting the polynomial as
P(x) = x⁴(1 − 4/x² − 1/x³ + 1/x⁴)
is to note that | −4/x² − 1/x³ + 1/x⁴ | < 1 for any sufficiently large |x| so
−1 < −4/x² − 1/x³ + 1/x⁴ < 1
and therefore
0 < 1 − 4/x² − 1/x³ + 1/x⁴ < 2
for any sufficiently large |x| so P(x) will be _positive_ for any sufficiently large |x|. And since P(−1) = −1, P(0) = 1, P(1) = −3 this means there will be at least one zero on each of the four intervals
(−∞, −1), (−1, 0), (0, 1), (1, ∞)
And since there can be no more than four zeros, the conclusion follows that P(x) has _exactly one zero_ on each of these four intervals.
Good thought, but that wouldn't really address the question of the roots. For example, for a quartic polynomial ( with leading coefficient positive ) to have 4 real roots ( as does the quartic that Mr. Prime Newtons considers here ) it is certainly NECESSARY that it also has 1 maximum and 2 minimums. But, conversely, that is NOT a sufficient condition for having 4 real roots. i.e. having 1 max and 2 mins does not guarantee 4 real roots, and in fact it does not guarantee any real roots ! To see this, imagine that the final numerical term in this quartic was the number 10 instead of 1. That would simply shift the graph UP by nine units. It would still have 1 max and 2 mins, but would not have any real roots.
Very Nice!
This is extremely complex, about 6 or 7 layers to this problem.
let say f(x)=x^2-5x+6 , x= 0, f(x)>0, is that mean f(x) doesnt cross x-axis? even q(x)>0 if x=0 in teh equation
No. The function you give can be written as
f(x) = (x − 2)(x − 3)
so we have f(2) = 0 and f(3) = 0. The graph of your function crosses the x-axis at x = 2 and at x = 3. The value of your function is negative for any real x between 2 and 3. You can graph your function online at the desmos website.
In fact this polynomial can be factored so it is octic solvable by radicals
(x^4 + 4x^2 + x + 1)(x^4 - 4x^2 - x + 1)
(x^2 - ax + b)(x^2 + ax + c) = x^4+4x^2+x+1
x^4 +ax^3+cx^2 - ax^3 - a^2x^2 - acx + bx^2+abx+bc = x^4+4x^2+x+1
x^4 + (b+c-a^2)x^2 + a(b-c)x + bc = x^4+4x^2+x+1
b+c-a^2 = 4
a(b-c) = 1
bc = 1
b+c = 4+a^2
b - c = 1/a
4bc = 4
2b = 4+a^2+1/a
2c = 4+a^2 - 1/a
4bc = 4
(4+a^2+1/a)(4+a^2 - 1/a) = 4
(4+a^2)^2-1/a^2 - 4 = 0
a^4+8a^2+16-4-1/a^2 = 0
a^4+8a^2+12-1/a^2=0
a^6+8a^4+12a^2-1=0
Here to find coefficients we must use cubic formula or derive it
For power sums there are also Newton - Girard formulas
so we can apply Newton - Girard formula and then Vieta formula
Too much work...
ESPECIALLY in a tournament where u only have such limited time
A root of an equation is the root of a sulotion to the equation, hm?
I'm a little puzzled, cause I might not have had this in math.
What is the point of taking a root of an equation?
Do you also take log or tan of an equation?
A root of an equation is a number that makes the expression equal 0.
@@xinpingdonohoe3978
Thank you very much! ❤️🙏
I need to take more time with this one to comprehend what the professor is doing.
Now I get it.
We call it "Nullstellen" (the spots, where the function is 0).
I still don't understand, why it is called "roots" in English. 🤔
@@kragiharproots of tree are seen at the ground, ground is x-axis?
@@lcex1649
Ahhhh.
Now I get it.
❤️🙏
Great 😊
Nice video
Bravo
Ouch!!! I intended Decade's
name to be spelled correctly!!!!
Descartes?
Be careful of autocorrect...
😮😮
asnwer=1 isit
(x ➖ 3x+1)
Wow!
The graphical arguments [and positive infinity and negative infinity] was awesome, but it involved a mature discussion, maybe not suited to younger members of your channel.
Genuinely, what are you talking about? Some strange patronisation to the уоungеr students is all you're providing.
@@xinpingdonohoe3978 It is not usual to relate solutions of a polynomial to concepts of calculus - it may not connect with the younger audience.
@@Arkapravo sure, if you're 6 it's not normal to do that, but if you're watching videos of Harvard-MIT tournaments you should be expecting *something* clever to occur.
@@xinpingdonohoe3978 Well put.
@@xinpingdonohoe3978 yeah, I am a fool
👍👍👍
Sturm’s method finds real roots.