A nice Math Olympiad Algebra Problem | Mathematical Olympiad Solutions
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- เผยแพร่เมื่อ 28 ก.พ. 2024
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In this Math Olympiad Algebra Equation, you'll learn tips and tricks for solving International Math Olympiad exams quickly. #IMO #matholympiad #algebra #radicalequations #simplify
Problem is solved partially by intuition.Are you solving it for an 8th std student? Please keep the standard of a '+2 student ' in the steps .
This one is specially solved for Olympiads
X=-4
I like your good thinking. I thought of using cubic equation solver with Cardano's formula but i did not notice your clever idea 80=16+64
Thanks 👍💯 for your support
Parece que la mejor resolución es la más artificial y extensa. Pero no es así. Este método sólo pudo seguirse porque se conocía a priori una solución, es decir, inútil y frustrante para el que mira el video queriendo aprender.
Esta ecuación se puede resolver en forma sencilla, por ejemplo nombrando z=1/x, la ecuación se reduce a
z^3- z^2+80=0, buscando raíces racionales se obtiene z=-4 , y dividiendo entre z+4, se obtienen las raices no reales. Luego se invierten para obtener x.
Si la ecuación no tuviera raíz racional
, podemos hacer un breve estudio con derivada de la función de 3er grado y aproximar sus raices. Si hay sólo una real repetimos el procedimiento anterior.
Varios problemas de matemática requieren estrategias complejas. No hay necesidad de mostrar complicado lo sencillo.
Thanks for your input
Nice equation. But the denominator for the quadratic formular is 2*A, Not 2*C
It's 2C as for that question. It's a new quadratic formula for solving equations of that form .
Let 1/X =a ,then a squared(1-a)=80,since a squared is positive, a must be negative and between 1 and 9.Its easy to see one correct answer immediately and to find the other two roots by synthetic division. Good Mathematics involves looking for intelligent shortcuts.Edricengert has the right idea.
I would instinctively look for powers of two and here you see that it’s 64 and 16 but since it’s only squared and cubed you know it has to be a power of four and given the negative sign it had to be -4 which means the denominator is -1/4. Hope I didn’t make a mistake!
(1/x)^2 - (1/x)^3 = 80
x ≠ 0
x/(x^3) - 1/(x^3) = 80
(x-1)/(x^3) = 80
x - 1 = 80*(x^3)
80*(x^3) - x + 1 = 0
solve this cubic equation we have:
x = -1/4
x = 1/8 + i(√55)/40
x = 1/8 - i(√55)/40
Nice solution. Thanks 👍
(1/x)²-(1/x)³=80
Domain
x≠0
Let y=(1/x)
y²-y³=80
y³-y²=-80
y³-y²+80=0
y³+64-y²+16=0
(y³+64)-(y²-16)=0
(y+4)(y²-4y+16)-(y+4)(y-4)=0
(y+4)(y²-5y+20)=0
y²-5y+20=0
4y²-20y+80=0
4y²-20y+25=-55
(2y-5)²=-55
|2y-5|=i√55
2y-5=±i√55
2(1/x)-5=±i√55
(2/x)-5=±i√55
(2/x)=5±i√55
x(5±i√55)=2
x=[2/(5±i√55)]
x=⅛±⅛i√2.2 ❤❤
y+4=0
y=-4
(1/x)=-4
-4x=1
x=-¼ ❤
Thanks 👍💯😊 for your input
Let's note X = 1/x. we have X^3 - X^2 +80 = 0. If a rational p/q (with p in Z, q in N*, p ^q = 1) is solution the p is a divisor of 80 and q is a divisor of 1 (then q=1), so the only possible rational solutions ara among the divisors of 80 in Z. We test... and finally -4 is OK.
Then X^3 - X^2 +80 = (X +4). (X^2 -5.X + 20) by polynomial division.
The roots of the second polynom are in C: (5 +i.sqrt(55)/2 and its conjuguatesqrt(, and with -4 they are the possibles values of X
Then the values of x are -1/4 ; 2/(5 +i.sqrt(55)) = (5 -i.sqrt(55))/40 and its conjuguate.
Nice solution. Thanks 👍💯 for your support
Also my idea. this is shorter way
I have a doubt why can't you directly take LCM on the first step and then you get an equation in the form of 80x^3 - x + 1 = 0
It's in the form of ax^3 + bx^2 + cx + d = 0 where b is 0 so it is not considered... Why can't you solve it that way?? Please enlighten me if I'm wrong.
Yes. It's possible but you'll have to multiply throughout the equation by 100 to make the first term a perfect cube. Thanks for your good question.
@@superacademy247 Ohh yea that makes sense but then It's a very lengthy process. Thank you, I really appreciate it.....
Hallo how are you its find i am graet you dimostration is good and clear tanks a gain
Hi. Thanks 👍💯😊 for your appreciation. Keep watching and learning new math skills
Sir, you make a small mistake that is not 2c it's 2a sir make a correction.Thank you sir
Check again. It's a special type of quadratic formula. Thanks for your concern though.
th-cam.com/video/PpcR6-OdrnM/w-d-xo.html
1. What does the world of mathematics benefit from artificial tasks that are designed to fit a good trick? Because if we change the task just a little, the trick no longer works. Like this:
(1/x)^2 - (1/x)^3 = 79
2. From around 3:30 onwards I would move a little faster and not indulge in numerous minimalist steps.
3. At 6:34 I would only take one step instead of three, namely:
1/x = -4
Changing to the reciprocal on both sides gives:
x = -1/4.
4. At 6:48 it looks like you're crossing out the equal sign.
5. At 7:30 it starts to really hurt. It takes a lot of time (four minutes!) and effort to calculate until the complex solutions are finally spit out. Instead, it's much more efficient like this:
5/x - 20 - 1/x^2 = 0
Multiply both sides by -x^2 and order in according to decreasing powers:
20x^2 - 5x + 1 = 0 => (x)2.3 = (5 +/- SQRT(24 - 80))/40 ...
6. From 15:05 you laboriously demonstrate the test for the real solution, even though you introduced the trick mentioned at the beginning with exactly the same calculation (4^2 + 4^3 = 80).
You have calculated the complex solutions, but the check for them is missing. And I noticed that at the beginning you didn't even mention the basic quantity in which the solution should be searched for: R or C ?
This question is awesome 💯👌😎. Look the introduction of another quadratic formula instead of 2a I've applied 2c and it works fine. If you concentrate throughout this solution you'll learn more than the tricks.
@@superacademy247Could it be that you wanted to lower your reply to cliveanawana5289 instead of to me?
About you remark #1: The pleasure to make a video show could be at the detriment of learning how to solve equations WHATEVER are the data values, unfortunately.
Mathematics is not about figuring out which trick to apply from a known set of tricks.
Mathematics is not just about knowing tricks or ready-made solutions, but rather about understanding the deeper principles that underlie various problems.
In the discussed problem, the matter is straightforward.
It can be seen that the substitution z=1/x has been made.
So, we substitute 1/x=z and obtain. z^2 - z^3 = 80.
We try to find rational solutions, which in this case must be integers:
z^2*(1-z) = 80.
It is evident that z^2 is positive, so 1-z is also positive, meaning z is less than 1, and z^2 divides 80, so we check z = -2... not enough.
z = -4... it is a solution, so now we are "smart" and know that 1/x = 1/4 will give the factorization with the factor (1/x - 1/4).
@@boguslawszostak1784:
Your first sentence is correct (the rest is unnecessary for me):
"Mathematics is not about figuring out which trick to apply from a known set of tricks."
That's exactly what you should tell the user "Super Academy", not me. Because he gave an example that was tailored to fit his solution trick. If you change the numbers just a little, the trick fails.
Why are you not letting 1/x be a or b for the quick solution?
You can if you wish to achieve quick result but in this video I wanted to demonstrate the authority of a new formula
Why it takes so much work to solve such simple problem?
The right hand side is an integer and is a small number,so the two terms on left hand side are small numbers. Power2 minus power 3 is positive means 1/X is negative. A few trials will get 1/X =-4. No pen and paper required
To demonstrate authority over the subject
Не проще ли заменить 1/х=t?
Yes! There are many methods
@@superacademy247Но ищут самый простой.А именно этот самый простой.
@@superacademy247 чем проще решать, тем меньше вероятность ошибки и тем меньше времени на решение. Особенно если это тесты
И каким образом из 1/х стал х на 10 минуте
By flipping over the equation on both sides. In other words, take the reciprocal on both sides
2c is wrong. It will be 2a
Its absolutely correct. It's a method applied to solve quadratic equations of that form. Watch and learn.
wrong application of quadratic root formulae.
It's a special type of quadratic formula used for solving reciprocal variables. I'll do a video that will highlight the strengths of the method.
On devise par 2a
Long route. U cant compete in olympiads. Sorry for the observation
This video is for educational purposes but in an exam setting you need to up your game and adjust your speed
(b2-4ac)/2a not (b2-4ac)/2c
It's a special type of quadratic formula applied to solve reciprocal variables. It's 2c because the formula defies traditional 2a