Solve Quartic Equation - Special Method | Chinese Math Olympiad

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How to solve this equation from the Mathematical Olympiad? This algebra video presents step-by-step the basic algebraic techniques and tricks to solve this quartic problem easily. For another Math Olympiad problem: • International Mathemat...
For more questions, tips and tricks like this, please visit Dr Wang's Channel: / @drwangusa
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27 окт 2023

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Комментарии : 49

@DrWangUSA 9 месяцев назад

Recommended Videos: Chinese Junior Math Olympiad: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-zQJYVHeCgSY.html International Math ContestTaiwan: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-kZH5fh1-3Dw.html Philippine Math Olympiad (Oral Competition): ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Xx3DWdu6TJw.html Singapore Math Olympiad : ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-cRaIK9OF4qc.html German Math Olympiad: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-B97HI0norzM.html

@damirdukic 8 месяцев назад

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 8 месяцев назад

Nice work! Thanks for sharing.

@avalagum7957 6 месяцев назад

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

@valvalrr 2 месяца назад

@@avalagum7957 damn you really thought you did something

@qwertyuiop2161 9 месяцев назад

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 9 месяцев назад

Same!

@DrWangUSA 9 месяцев назад

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

@DrWangUSA 9 месяцев назад

Here is the question in our one of our videos solved using the method you mentioned: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-xAitvHF81J8.html

@nasrullahhusnan2289 8 месяцев назад

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

@DrWangUSA 8 месяцев назад

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

@nemesiochupaca5024 6 месяцев назад

@@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.

@tejpalsingh366 3 месяца назад

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

@herbertklumpp2969 8 месяцев назад

(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 8 месяцев назад

Great work and thanks for sharing!

@user-mf5nw9qg5r 8 месяцев назад

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 8 месяцев назад

Nice work and thanks for sharing!

@ZhilinChen-my7tp Месяц назад

a+b is not 1

@user-mf5nw9qg5r Месяц назад

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

@hugh081 7 месяцев назад

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 7 месяцев назад

Nice method and thanks for sharing

@tunneloflight 8 месяцев назад

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 8 месяцев назад

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 8 месяцев назад

@@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 8 месяцев назад

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.

@andrec.2935 7 месяцев назад

Linda solução!

@DrWangUSA 7 месяцев назад

Thanks

@user-xm6ev1tv1i 7 месяцев назад

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

@DrWangUSA 7 месяцев назад

Thanks for sharing

@MasterOfCubes2074 8 месяцев назад

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 8 месяцев назад

Nice work and thanks for sharing!

@prollysine 8 месяцев назад

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 8 месяцев назад

Nice work and thanks for sharing!

@prollysine 8 месяцев назад

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

@markandeysingh1355 5 дней назад

There should be Four Values of x.

@nicholasngo5428 7 месяцев назад

x=0

@jim2376 7 месяцев назад

By inspection x = 0 works.

@DrWangUSA 7 месяцев назад

Nice work

@user-nd7th3hy4l 7 месяцев назад

X=0 est trivial. On verra le reste.

@user-nd7th3hy4l Месяц назад

X=0

@LuisSayago-ec3hq 2 месяца назад

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 2 месяца назад

Hol up

@user-it6fh7hy6t 7 месяцев назад

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 7 месяцев назад

Thanks for sharing