Solve Quartic Equation - Special Method | Chinese Math Olympiad

[@DrWangUSA]

Recommended Videos: Chinese Junior Math Olympiad: kzbin.info/www/bejne/sIKtiol-mqiaibs International Math ContestTaiwan: kzbin.info/www/bejne/oYurZpmeZpJmetk Philippine Math Olympiad (Oral Competition): kzbin.info/www/bejne/jqmWdYqaqpuHgNk Singapore Math Olympiad : kzbin.info/www/bejne/mYPEen5vhKtnp8U German Math Olympiad: kzbin.info/www/bejne/eGqaeXxmo9SlsK8

[@qwertyuiop2161]

My method was to introduce symmetry by subistution u=x+3/2. This creates a biquadratic when expanded which can be easily solved. then solve back for x.

[@actions-speak]

Same!

[@DrWangUSA]

Thanks for sharing! Symmetry idea is also an important method to solve high degree equations to cancel odd terms.

[@DrWangUSA]

Here is the question in our one of our videos solved using the method you mentioned: kzbin.info/www/bejne/rnLMpal-e51kgJo

[@damirdukic]

Well, I first immediately saw that "0" was a solution. _(Because 17 = 2^4 + 1^4)._ Then I noticed that there should have been at least one negative solution _(because (-a)^4 = a^4),_ and after a little pondering I saw that that negative solution was "-3". _(Because 1 - 3 = -2 and 2 - 3 = -1.)_ After that, I simply did the expansion (which is actually _not so difficult_ to do here), then I divided the resulting equation first with "2x" and then the result of that division with "x+3". The resulting quadratic equation "x^2 + 3x + 6 = 0" doesn't have any real solutions, so "0" and "-3" are the only real solutions of the initial equation.

[@DrWangUSA]

Nice work! Thanks for sharing.

[@avalagum7957]

This solution is simpler and more intuitive than the one in the video.

[@valvalrr]

@@avalagum7957 damn you really thought you did something

[@鈞齊]

Recall: a⁴+b⁴+(a+b)⁴=2(a²+ab+b²)² (x+1)⁴+(x+2)⁴=17 => 1⁴+(x+1)⁴+(x+2)⁴=18 => (t²+t+1)²=9, where t=x+1 => t²+t+4>0 , for all t in R t²+t-2=0 => t=1 or -2 => x=0 or -3

[@DrWangUSA]

Nice work and thanks for sharing!

[@ZhilinChen-my7tp]

a+b is not 1

[@鈞齊]

@@ZhilinChen-my7tp 1+(x+1)=x+2

[@nasrullahhusnan2289]

Note that 17=16+1 =2⁴+1⁴ (x+2)⁴+(x+1)⁴=17 By inspection it is clear that x=0

[@DrWangUSA]

Nice work! X=-3 is also an integer solution. You may need to show that there are no other real number solutions.

[@nemesiochupaca5024]

@@DrWangUSA With the same observation that the youtuber made, given that it is a quartic, the cases must be considered: (-(x+1))^4 = 2^4 ⇒ -x-1=2 ⇒ x=-3 (-(x+2))^4 = 1^4 ⇒ -x-2=1 ⇒ x=-3 And the other integer solution -3 is also obtained.

[@herbertklumpp2969]

(Y+1)^4 +y^2 = 17 conclude Y^4 +2y^3 +3y^2 +2y -8=0 Y=0 y=-2 are solutions factorizing you get (Y-1)*(Y+2) ( y^2+y+4) =0 the last one has no real solution therefore y=1 or y=-2 conclude x=0 x=-3

[@DrWangUSA]

Great work and thanks for sharing!

[@tunneloflight]

Easy. The two terms must be perfect 4th powers. The only numbers satisfying that are 2 and 1. 2^4+1^4 =17. So what values of x result in 1 and 2 inside the 4th powers. (X+1), (X+2). Easy = X=0, and X=-3.

[@DrWangUSA]

Great idea. From given condition, x is a real number. You derived the integer solutions. You may need to show that there is no other real situations.

[@tunneloflight]

@@DrWangUSA you are right. As a quartic, there are potentially two other solutions. As two smooth solutions, the two parts sum will be the sum of two shapes that rise symmetrically with increasingly positive or negative values for X+1 and X+2. They are constrained to be within the range of -3 to 0. These would need to be fractions that are themselves perfect 4th power sums. The constraints on that likely rule out a solution. Though I don’t immediately see the constraints that do rule out all solutions. Interesting.

[@tunneloflight]

Ah, I see it now. The shape of the curves dictates that the value of the sum will decline as the value of X approaches the two poles at (1, 2), i.e. X values of (-1, -2). And the sum will continue to decline as X approaches the midpoint at -1.5. As a consequence, no real solutions can exist either between the two main solutions (-3, 0), nor at values greater than 1, nor less than -3. The non-imaginary solutions are then constrained to 0 and -3.

[@MasterOfCubes2074]

Let x+1=t Hence, x+2=x+1+1=t+1 (t+1)⁴+t⁴=17 (t+1)⁴+t⁴-17=0 [(t+1)²]²+t⁴-17=0 [t²+2t+1]²+t⁴-17=0 t⁴+4t³+6t²+4t+1+t⁴-17=0 2t⁴+4t³+6t²+4t-16=0 By factorization 2(t−1)(t+2)(t²+t+4)=0 2(t-1)=0, t=1 (t+2)=0, t=-2 (t²+t+4)=0 a=1 b=1 c=4 D=b²-4ac D=1-4×4×1 D=1-16 D=-15 x=t x = (-b ± √ (D) )/2a t=-1±√( -15)/2 t=-1±i√(15)/2 t is either equal to -(1+i√(15))/2or i√(15)-1/2. We have now 4 roots of this equation t=1 t=-2 t=-(1+i√(15))/2 t=i√(15)-1/2 We know that x+1=t x+1=1 x=0 x+1=-2 x=-3 x+1=-(1+i√(15))/2 x=(-(1+i√(15))/2)-1 x=-(3+i√(15))/2 x+1=i√(15)-1/2 x=(i√(15)-1/2)-1 x=i√(15)-3/2 I think this is esaier method. As this is 4th power equation hence it have 4 roots but in question value of x ε R Hence, according to question the value if x is either 1 or -3.

[@DrWangUSA]

Nice work and thanks for sharing!

[@prollysine]

Sir, Hi, we get a solution by expanding the powers: 2x^4+12x^3+30x^2+36x+17=17 -> x(x^3+6x^2+15x+18)=0 , >>x=0 x^3+3x^2 + 3x^2+9x + 6x+18=0 -> (x+3)(x^2+3x+6)=0 , x+3=0 , >>x=-3

[@DrWangUSA]

Nice work and thanks for sharing!

[@prollysine]

Dr. Wang, thank you for appreciating my paper, it presents an interesting example, regards...@@DrWangUSA

[@hugh081]

Expand and combine terms, and the equation is divisible by x, so x=0 is a solution. The remaining cubic is fairly easy to solve

[@DrWangUSA]

Nice method and thanks for sharing

[@바다노을-o3r]

16+1=17 2^4 + 1^4 = 17 (-1)^4 + (-2)^4 =17 따라서 x=0 또는 x=-3

[@DrWangUSA]

Thanks for sharing

[@Quest3669]

Its longer method ...solve just by making it a+b whole square form ....thts easy n wuick

[@markandeysingh1355]

There should be Four Values of x.

[@andrec.2935]

Linda solução!

[@DrWangUSA]

Thanks

[@LuisSayago-ec3hq]

My solution is TRIVIAL..Making Y1=(X+2)^4 & Y2=(X+1)^4 then Y1=Y2 & By finding the values for both equations, we have that for Y1, (X,Y)=(0,16); (-1,1); (-2,0); (-3,1), etc & for Y2, (X,Y)=(0,16); (-1,17); (-3,1) etc.then the common values in both results are (0,16); (-3,1).. therefore the solutions of this question are X=0 ; X=-3...GOT IT??

[@valvalrr]

Hol up

[@ЭдуардПлоткин-р3л]

Зачем же так сложно решать такое легкое уравнение? (x+2)⁴-16+(x+1)⁴-1=0 Дальше используем разность квадратов и решение укладывается в 4 минуты. 2x(x+3)(x²+3x+6)=0 На вопросы не отвечу,нет свободного времени.

[@DrWangUSA]

Thanks for sharing

[@jim2376]

By inspection x = 0 works.

[@DrWangUSA]

Nice work

[@Bertin-q3y]

X=0 est trivial. On verra le reste.

[@nicholasngo5428]

x=0

[@Bertin-q3y]

X=0