The trick of solving palindromic quartics can also be applied to general quartics of the form ax^4 + bx^3 + cx^2 + dx^ + e = 0 provided this condition is satisfied => (a/e) = (b/d)^2
Well, my recommendation is to do the "double cross factorization": First we list two rows as following with each row three numbers. a1 b1 c1 × × a2 b2 c2 Then find numbers that match these equations: a1*a2=2(coefficient of power 4); c1*c2=18(coefficient of power 0); a1*b2+a2*b1=7(coefficient of power 3); b1*c2+b2*c1=-21(coefficient of power 1); a1*c2+a2*c1+b1*b2=-34(coefficient of power 2); Now we get the double cross array like this: 1 3 -18 × × 2 1 -1 Convey it directly into factors: (x²+3x-18)(2x²+x-1)=0 Do further factorizing for these two seperate power 2 polies. (x-3)(x+6)(x+1)(2x-1)=0 Now each of the 4 roots can be found.
The reason the so-called palindromic method works with this non-palindromic equation is that this actually _is_ an almost palindromic equation but _in disguise._ To see this, substitute x = √3·u then we have 18u⁴ + 21√3·u³ − 102u² − 21√3·u + 18 = 0 So, the equation actually is palindromic except for the opposite signs of the coefficients of the cubic and the linear term, but these near palindromic quartics can still be solved similarly to the usual solution method for true palindromics. If we divide both sides by u² and group terms with equal or opposite coefficients we have 18(u² + u⁻²) + 21√3·(u − u⁻¹) − 102 = 0 and with a substitution u − u⁻¹ = s and therefore u² + u⁻² = s² + 2 this becomes 18(s² + 2) + 21√3·s − 102 = 0 18s² + 21√3·s − 66 = 0 To rationalize this quadratic we can substitute s = t/√3 which gives 6t² + 21t − 66 = 0 2t² + 7t − 22 = 0 which is the exactly the quadratic in t obtained in the video. It is easy to see why. We have s = t/√3 so t = √3·s and s = u − u⁻¹ and x = √3·u so u = x/√3 which gives t = √3·s = √3·(u − u⁻¹) = √3·(x/√3 − √3/x) = x − 3/x and of course t = x − 3/x is exactly the substitution used in the video.
Pretty nice & creative way of factorizing.Well, my recommendation is to do the "double cross factorization": First we list two rows as following with each row three numbers. a1 b1 c1 × × a2 b2 c2 Then find numbers that match these equations: a1*a2=2(coefficient of power 4); c1*c2=18(coefficient of power 0); a1*b2+a2*b1=7(coefficient of power 3); b1*c2+b2*c1=-21(coefficient of power 1); a1*c2+a2*c1+b1*b2=-34(coefficient of power 2); Now we get the double cross array like this: 1 3 -18 × × 2 1 -1 Convey it directly into factors: (x²+3x-18)(2x²+x-1)=0 Do further factorizing for these two seperate power 2 polies. (x-3)(x+6)(x+1)(2x-1)=0 Now each of the 4 roots can be found.
I like the method (but I'm not sure how generally applicable it is). By comparison, I immediately took the hints from the equation that (2x±1) and (x±3) and (x±1) could be solutions. That gave me (checking, long division) (2x-1)(x-3)(x+1)(x+6). It didn't take long.
2x^4+7x^3-34x^2-21x+18 = 0 16x^4+56x^3-272x^2-168x+144 = 0 (16x^4+56x^3) - (272x^2+168x-144)=0 (16x^4+56x^3+49x^2) - (321x^2+168x-144) = 0 (4x^2+7x)^2 - (321x^2+168x-144) = 0 (4x^2+7x+y/2)^2 - ((4y+321)x^2+(7y+168)x+y^2/4-144) = 0 Here without calculating disriminant we know that we can put y = -24 but in general we have to calculate discriminant of the quadratic and equate it to zero 4(y^2/4-144)(4y+321)-(7y+168)^2=0 (4x^2+7x-12)^2 - (-96+321)x^2= 0 (4x^2+7x-12)^2 - 225x^2= 0 ((4x^2+7x-12) - 15x)((4x^2+7x-12) + 15x) = 0 (4x^2-8x-12)(4x^2+22x-12) = 0 (x^2 - 2x - 3)(2x^2 + 11x - 6) = 0
The polynomial still satisfies the condition that if x is a root there exists a k such that k/x is also a root so the method is the same as the one for palindromic quartics (where k=1). Such a k exists iff a.d^2 = e.b^2.
sry for my bad english! 1:02 guessing isn´t a good solove i believe there have to be a formular! 2:41 make sence 5:49 looks like simplify it 6:20 ok and now the abc-formular 7:23 that i don´t understood but ok 9:00 so we have 4 soulutions? 10:30 nice sentence see you K.Furry
sir please could you solve this jee advanced 2016 question? Paper 2: Maths Section 2 :- Question 44 ( multiple options correct ) its a limit question average time per question is 3-4 mins in some places its given as question 44 also
Turn each product of a_n into e^sum of ln a_n. Then it becomes a Riemann sum, so you have an integral from 0 to 1. The first is x ln(1+ux) du, and the second is -x ln(1+u²x²) du. Substitute t=ux. Then differentiate, using FTC, and observe that it's increasing for (0,1).
The conventional approach still works. Guess that the answer is of the form (x^2 + ax + 6)(2x^2 + bx + 3). Expand it out. If you can pick values a and b, you have a solution. If not, then try making the 3 and 6 negative. The other thing to try is swapping the 1 and the 2. In this case, you end up with (x^2 - 2x - 3)(2x^2 + 11x - 6). Which are the two quadratics he got in the video, so that's a good sign.
I dont think that method you presented works for every quartic equation It depends on that what coefficients you have Even if you find method for general quartic which is based on method for palindromic equations you will not be able to avoid cubic resolvent
Well, most of these equations in exams are supposed to have rational roots to make solving easy, or a convenient substitution. Obviously a general quartic equation can only be solved properly using the very long unmemorizable formulas, because for instance equations with 2 irrational roots and 2 complex ones aren't gonna be found out easily.
@@niloneto1608 Not necessarily. If the quartic factors into two quadratics with integer coefficients (as is usually the case with quartics that show up on math contests) _but_ these quadratics both have irrational or complex roots, then it is still perfectly possible to solve these quartics _without_ memorizing long formulas, using e.g. Ferrari's method. Sure, you will get a cubic resolvent, but that will then have a rational solution which is either an integer or an integer multiple of ½ and you don't even have to use the rational root theorem to solve that cubic resolvent. See my main comment on this video where I solve the quartic from this video using Ferrari's method.
Hi sir, could you make a video(s) about graph theory and how to apply it to problems like the AMC,AIME, or other problems. I ask you because I saw problem 10 of 2019 AMC 12B. Thanks.
This has nothing to do with π. Generally, for any two positive real numbers a and b you have aˡⁿ ᵇ = bˡⁿ ᵃ You will see immediately why this is so if you consider that a = eˡⁿ ᵃ b = eˡⁿ ᵇ and (xᵐ)ⁿ = xᵐⁿ = xⁿᵐ = (xⁿ)ᵐ for any positive real x and any real m and n, so you have aˡⁿ ᵇ = (eˡⁿ ᵃ)ˡⁿ ᵇ = (eˡⁿ ᵇ)ˡⁿ ᵃ = bˡⁿ ᵃ
x^ln(π) can rewritten in terms of base e as (e^ln(x))^ln(π). If you multiply the exponents as e^(ln(x)lnπ)), we can now use properties of logarithms as e^(ln(π^ln(x)))=π^ln(x). Remember, you can interchange the base and the log of argument as long as you keep the base of the log the same meaning that a^(log_b(c))=c^(log_b(a)).
I have the formulas for computing the four roots symbolically. I had a problem where the solution for time was a quartic. I only needed the positive real root. What happens if I change the equation in this video. Is the solution general?
@@pnachtwey Sure, but ready-made formulas to express the roots of a quartic in terms of its coefficients are complicated and therefore not suitable for application on e.g. math contests, where it is an advantage if you can easily reduce the solution of a quartic to the solution of a quadratic. But there are also procedures like Ferrari's method which _are_ applicable to any quartic equation and which are easy to remember. See my main comment where I solve the quartic equation from the video using Ferrari's method.
@@NadiehFan The math tests are just trick tests then and not how to apply the tricks. I needed to find the roots of a quartic equation for motion control. The roots should have units of time. Two of the roots are real and two are complex. If the two real roots are positive, then extra work had to be done to figure out which one to use. Otherwise, I only needed the positive real root since time can't be imaginary or negative. Tricks rarely if ever work in real world applications so why bother?
The trick of solving palindromic quartics can also be applied to general quartics of the form ax^4 + bx^3 + cx^2 + dx^ + e = 0 provided this condition is satisfied => (a/e) = (b/d)^2
my "double cross" method solves every general quartics as the form ax⁴+bx³+cx²+dx+e=0, without any condition.
Well, my recommendation is to do the "double cross factorization":
First we list two rows as following with each row three numbers.
a1 b1 c1
× ×
a2 b2 c2
Then find numbers that match these equations:
a1*a2=2(coefficient of power 4);
c1*c2=18(coefficient of power 0);
a1*b2+a2*b1=7(coefficient of power 3);
b1*c2+b2*c1=-21(coefficient of power 1);
a1*c2+a2*c1+b1*b2=-34(coefficient of power 2);
Now we get the double cross array like this:
1 3 -18
× ×
2 1 -1
Convey it directly into factors:
(x²+3x-18)(2x²+x-1)=0
Do further factorizing for these two seperate power 2 polies.
(x-3)(x+6)(x+1)(2x-1)=0
Now each of the 4 roots can be found.
The reason the so-called palindromic method works with this non-palindromic equation is that this actually _is_ an almost palindromic equation but _in disguise._ To see this, substitute
x = √3·u
then we have
18u⁴ + 21√3·u³ − 102u² − 21√3·u + 18 = 0
So, the equation actually is palindromic except for the opposite signs of the coefficients of the cubic and the linear term, but these near palindromic quartics can still be solved similarly to the usual solution method for true palindromics. If we divide both sides by u² and group terms with equal or opposite coefficients we have
18(u² + u⁻²) + 21√3·(u − u⁻¹) − 102 = 0
and with a substitution u − u⁻¹ = s and therefore u² + u⁻² = s² + 2 this becomes
18(s² + 2) + 21√3·s − 102 = 0
18s² + 21√3·s − 66 = 0
To rationalize this quadratic we can substitute s = t/√3 which gives
6t² + 21t − 66 = 0
2t² + 7t − 22 = 0
which is the exactly the quadratic in t obtained in the video. It is easy to see why. We have s = t/√3 so t = √3·s and s = u − u⁻¹ and x = √3·u so u = x/√3 which gives
t = √3·s = √3·(u − u⁻¹) = √3·(x/√3 − √3/x) = x − 3/x
and of course t = x − 3/x is exactly the substitution used in the video.
Pretty nice & creative way of factorizing.Well, my recommendation is to do the "double cross factorization":
First we list two rows as following with each row three numbers.
a1 b1 c1
× ×
a2 b2 c2
Then find numbers that match these equations:
a1*a2=2(coefficient of power 4);
c1*c2=18(coefficient of power 0);
a1*b2+a2*b1=7(coefficient of power 3);
b1*c2+b2*c1=-21(coefficient of power 1);
a1*c2+a2*c1+b1*b2=-34(coefficient of power 2);
Now we get the double cross array like this:
1 3 -18
× ×
2 1 -1
Convey it directly into factors:
(x²+3x-18)(2x²+x-1)=0
Do further factorizing for these two seperate power 2 polies.
(x-3)(x+6)(x+1)(2x-1)=0
Now each of the 4 roots can be found.
this is a pretty sick method
@@mentooo5709 it works well,best thing is you don't need ananysis to the polynomial
I like the method (but I'm not sure how generally applicable it is).
By comparison, I immediately took the hints from the equation that (2x±1) and (x±3) and (x±1) could be solutions.
That gave me (checking, long division) (2x-1)(x-3)(x+1)(x+6). It didn't take long.
Awesome. Loved it.
As it turns out, the Rational Roots theorem solved this equation completely
But of course. That's like using the QF on a QE and after obtaining integer solutions admitting that it could have been factored 😀
The guy says he thinks his method is faster but I think RRT is much faster
2x^4+7x^3-34x^2-21x+18 = 0
16x^4+56x^3-272x^2-168x+144 = 0
(16x^4+56x^3) - (272x^2+168x-144)=0
(16x^4+56x^3+49x^2) - (321x^2+168x-144) = 0
(4x^2+7x)^2 - (321x^2+168x-144) = 0
(4x^2+7x+y/2)^2 - ((4y+321)x^2+(7y+168)x+y^2/4-144) = 0
Here without calculating disriminant we know that we can put y = -24
but in general we have to calculate discriminant of the quadratic and equate it to zero
4(y^2/4-144)(4y+321)-(7y+168)^2=0
(4x^2+7x-12)^2 - (-96+321)x^2= 0
(4x^2+7x-12)^2 - 225x^2= 0
((4x^2+7x-12) - 15x)((4x^2+7x-12) + 15x) = 0
(4x^2-8x-12)(4x^2+22x-12) = 0
(x^2 - 2x - 3)(2x^2 + 11x - 6) = 0
The polynomial still satisfies the condition that if x is a root there exists a k such that k/x is also a root so the method is the same as the one for palindromic quartics (where k=1). Such a k exists iff a.d^2 = e.b^2.
Never received a notification this early from UA-cam.
I need more similar question like this one
sry for my bad english!
1:02 guessing isn´t a good solove i believe there have to be a formular!
2:41 make sence
5:49 looks like simplify it
6:20 ok and now the abc-formular
7:23 that i don´t understood but ok
9:00 so we have 4 soulutions?
10:30 nice sentence
see you
K.Furry
Very useful!
sir please could you solve this jee advanced 2016 question?
Paper 2: Maths Section 2 :- Question 44 ( multiple options correct )
its a limit question
average time per question is 3-4 mins
in some places its given as question 44 also
can you write it here??
@@savitatawade2403 i dont know how to type those symbols but lemme try
@@djzodiac9075 yes please
Turn each product of a_n into e^sum of ln a_n. Then it becomes a Riemann sum, so you have an integral from 0 to 1. The first is x ln(1+ux) du, and the second is -x ln(1+u²x²) du.
Substitute t=ux. Then differentiate, using FTC, and observe that it's increasing for (0,1).
@@savitatawade2403 Yt keeps thinking it’s a spam comment ands keep removing it
I didnt know how to do this on paper Thank u for making me know
The conventional approach still works. Guess that the answer is of the form (x^2 + ax + 6)(2x^2 + bx + 3). Expand it out. If you can pick values a and b, you have a solution. If not, then try making the 3 and 6 negative. The other thing to try is swapping the 1 and the 2. In this case, you end up with (x^2 - 2x - 3)(2x^2 + 11x - 6). Which are the two quadratics he got in the video, so that's a good sign.
Thanks sir
I dont think that method you presented works for every quartic equation
It depends on that what coefficients you have
Even if you find method for general quartic which is based on method for palindromic equations
you will not be able to avoid cubic resolvent
Well, most of these equations in exams are supposed to have rational roots to make solving easy, or a convenient substitution. Obviously a general quartic equation can only be solved properly using the very long unmemorizable formulas, because for instance equations with 2 irrational roots and 2 complex ones aren't gonna be found out easily.
@@niloneto1608 Not necessarily. If the quartic factors into two quadratics with integer coefficients (as is usually the case with quartics that show up on math contests) _but_ these quadratics both have irrational or complex roots, then it is still perfectly possible to solve these quartics _without_ memorizing long formulas, using e.g. Ferrari's method. Sure, you will get a cubic resolvent, but that will then have a rational solution which is either an integer or an integer multiple of ½ and you don't even have to use the rational root theorem to solve that cubic resolvent. See my main comment on this video where I solve the quartic from this video using Ferrari's method.
Hi sir, could you make a video(s) about graph theory and how to apply it to problems like the AMC,AIME, or other problems. I ask you because I saw problem 10 of 2019 AMC 12B. Thanks.
I would like to know more about you such as where you studied, etc. Do you have a Linkedin page?
So does this always work for quartic equations?
No.
Sir kindly make video on vector space
If we put -1 we can see it's a factor from which we can easily find the cubic eqn😅
Solving some inequalities please
Hi please can you explain why is x to the power natural log of Pi is equal to pi to the power natural log of x
This has nothing to do with π. Generally, for any two positive real numbers a and b you have
aˡⁿ ᵇ = bˡⁿ ᵃ
You will see immediately why this is so if you consider that
a = eˡⁿ ᵃ
b = eˡⁿ ᵇ
and
(xᵐ)ⁿ = xᵐⁿ = xⁿᵐ = (xⁿ)ᵐ
for any positive real x and any real m and n, so you have
aˡⁿ ᵇ = (eˡⁿ ᵃ)ˡⁿ ᵇ = (eˡⁿ ᵇ)ˡⁿ ᵃ = bˡⁿ ᵃ
x^ln(π) can rewritten in terms of base e as (e^ln(x))^ln(π). If you multiply the exponents as e^(ln(x)lnπ)), we can now use properties of logarithms as e^(ln(π^ln(x)))=π^ln(x). Remember, you can interchange the base and the log of argument as long as you keep the base of the log the same meaning that a^(log_b(c))=c^(log_b(a)).
why not just use the Horner's scheme🤨
isnt polynomdivision not possible ?
14 mins and 14 likes nice!
Edit 17 mins 17 likes and 23mins 23 like!! Interesting pattern
59 mins and 59 likes! This cannot be a coincidence
@@neevhingrajia3822420 minutes, 201 likes, yes it can
2x⁴+7x³-34x²-21x+18=0
(2x²-7x+3)(x²+7x+6)
(2x²-7x+3)=0
(2x -1)(x-3)=0
x= 1/2, x= 3 #
(x²+7x+6)=0
(x+1)(x+6)=0
x= -1, x= -6 #
i made this by myself
Hasn't any woke weirdos told this gentleman yet?
Math is RACIST 😮
Wtf? Why bringing cultural stuff here where it doesn't matter a single bit?
@@niloneto1608 Sorry but you can't hide from it forever.
I have the formulas for computing the four roots symbolically. I had a problem where the solution for time was a quartic. I only needed the positive real root. What happens if I change the equation in this video. Is the solution general?
No, this is not a method that can be applied to any quartic equation.
@@NadiehFan then why bother? My equations for solving quartics ARE GENERAL!
@@pnachtwey Sure, but ready-made formulas to express the roots of a quartic in terms of its coefficients are complicated and therefore not suitable for application on e.g. math contests, where it is an advantage if you can easily reduce the solution of a quartic to the solution of a quadratic. But there are also procedures like Ferrari's method which _are_ applicable to any quartic equation and which are easy to remember. See my main comment where I solve the quartic equation from the video using Ferrari's method.
@@NadiehFan The math tests are just trick tests then and not how to apply the tricks. I needed to find the roots of a quartic equation for motion control. The roots should have units of time. Two of the roots are real and two are complex. If the two real roots are positive, then extra work had to be done to figure out which one to use. Otherwise, I only needed the positive real root since time can't be imaginary or negative.
Tricks rarely if ever work in real world applications so why bother?