Hello everyone, welcome to my channel. I share how to solve Math Olympiad questions easily and quickly. I share information about examples of Olympic questions and tricks for solving them. Discusses exponents, algebra, mathematical roots, simplifying roots, multiplication, logarithms, and basic mathematics. Happy watching and always happy. Thank you to all of you💖
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no way this is a problem from a math olympiad, it's basic
Ofc the answer has to be in the complex plane, as no real solution can exist.
Nice work 👏
Thankyou
Потеряны ещё 2 корня для a. 1. a=(1+√533)/2 2. a=(1-√533)/2 Решение. Имеем (1) a²=b+133 (2) b²=a+133 (1) b=a²-133 (2) a²-133 = a+133 a²-133 = a+133 --> a⁴-266a²-a+17556=0 Два корня найдены для а. Это 11 и -12. Делим многочлен сначала на (а-11), затем на (a+12). Получаем a² - a - 133. То есть имеем уравнение (a-11)(a+12)(a² - a - 133)=0 Нужно решить ещё одно уравнение a² - a - 133 = 0, откуда a=(1+√533)/2; a=(1-√533)/2. b=a²-133
Я забыл скобки написать. Должно быть (a²-133)² = a+133
Is this really the Olympic question
This can be solved by observation. At glance you can see that 36 is a sum of 27 and 9. Given powers, m must be 3^1/3
You were so close. You've got a mistake in your last step. sqrt(4 sqrt(2) + 4) cannot be simplified to 2 sqrt(2) + 2. The former is ~3.1, which is the correct answer, but the latter is ~4.8. Anyway, this is a pretty simple system of equations with two variables, so it's easily solved for a and b by saying b = 2/a and plugging that back into the first equation and solving for the quadratic in a^2. a = sqrt(2 sqrt(2) + 2) and b = sqrt(2 sqrt(2) - 2).
If you have to do all that to respond to that question, you'll never have time to respond to the other questions of the Olympiad. The answer was trivial from the beginning.
:(
I know I am not too smart but Honestly I figured it out in 10 seconds after seeing the thumbnail.😂
hi my name is Ayush chaurasia actually i am from India. Been a early mathematician i have discovered many things in the field of maths.which indludes a best way to solve this problem by a formula.which is just ( S±√S2 + 4P ÷ 2 ).
Hi Ayush, nice to meet you, thank you for the information, I will learn about this easy method😊
1,5
Nice🥰
1,5.
Nice😍
@@Ms_Math1 Thanks!
Let x=a+b and y=a-b. x+y=(a+b)+(a-b)=2a=6, thus a=3. xy=(a+b)(a-b)=a^2-b^2=3^2-b^2=6, thus b^2=3, or b=+-sqrt(3).
Once we get to 4^x=8 we can take a much more simpler path : 4^x=8 (2^2)^x=2^3 2^(2x)=2^3 2x=3 x=3/2
Не 2/3, а 3/2, т. к. делим на коэффициэнт при неизвестном
@@user-ec5ip3vp2r oh yes thank you for pointing out my typo
4^x+4^x=16 2^2x+2^2x=16 2*2^2x=16 2^2x=8 2^2x=2^3 2x=3 x=3/2
😮
2^ (x+1) = 2^4 --> x=2/3 (Okham)
How about taking log early?
In minute 2:56 why 24 change to 16?
Sorry, that's not 24 but 2 to the power of 4. sorry if my writing isn't clear to read
without any variable (x): (a^¼) + (√a) = 1 ... 🟣 → (a^¼) = (1 - √a) ... 🔴 [ (a^¼) + (√a) ]² = 1² ... ( from 🟣² ) → (√a) + (a) + 2(a^¼)(√a) = 1 → (√a) + (a) + 2(1 - √a)(√a) = 1 ... ( from 🔴) → (√a) + (a) + (2√a) - (2a) = 1 → (-a) + (3√a) = 1 ... 🔵 → (√a) = ⅓ (a + 1) ... 🟢 [ (-a) + (3√a) ]² = 1² ... ( from 🔵² ) → (a²) + (9a) - 6a(√a) = 1 → (a²) + (9a) - 6a (⅓ (a+1) ) = 1 ... ( from 🟢) → (a²) + (9a) - 2a(a+1) = 1 → (a²) + (9a) - (2a²) - 2a = 1 → -(a²) + (7a) = 1 → (a²) - 7a + 1 = 0 → a = [ 7 ± √(49-4) ] / 2 → a = ½ ( 7 ± √45 ) → a = ½ [ 7 ± √(9*5) ] → a = ½ ( 7 ± 3√5 ) eliminate ½ ( 7 + 3√5 ) because (7/2) > 1. → a = ½( 7 - 3√5 )
8^x+2^x=30 2^(3x)+2^x-30=0 t=2^x, t>0 t³+t-30=0 f(t)=t³+t-30 f(3)=27+3-30=0 f(t)=t³-3t²+3t²-9t+10t-30=t²(t-3)+3t(t-3)+10(t-3) f(t)=(t-3)(t²+3t+10) t-3=0 or t²+3t+10=0 t=3 or t²+3t+2.25=-7.75 2^x=3 or (t+1.5)²=-7.75 x=log2(3) or x is not real
3
1080p, but i can't see anything
(m-n)^2=144 m^2+n^2=144+2mn m^2+n^2+2mn=144+4mn m+n= (144+4mn)^(1/2)
(-12;11)(11;-12)
(-16;15)(15;-16)
x=1+z and y=-1+z; (z+1)(z-1)=z^2-1=2; z=±√3; x=1+z=1±√3 and y=-1+z=-1±√3
64-25=39; a=±8 and b=±5
m=z+6 and n=z-6 (z+6)(z-6)=z^2-36=12 or z=±4√3 m+n=2z=±8√3 Check: m=6±4√3 and n=-6±4√3; m=6±4√3=n+12=-6±4√3+12; mn=(6±4√3)(-6±4√3)=-36+48=12
The solution for this problem is the following: X=7 & Y=5. 7 to the 2nd power is 49 and 5 to the 2nd power is 25. 49-25=24, 7*5=35, and 7+5=12. It's as simple as all that.
basically, I observed that p^2 > p^3. It leads me to think that p must be < 0. Next, the disparity lies in integers, and my hunch suggests that this number is highly probable to be an integer. The cubic of this number must be closer to 80. So, I take -4, (-4)^3 = -64. I am almost to guess this number to satisfy this equation. Does my thinking process make any mistake here?
I did that also.
answer for any a,b : if x^2+y^2 = a and xy=b x+y = ±sqrt(±sqrt(4b^2+a^2)+2b))
An efficient way to solve x² − y² = 24 xy = 35 for x + y is to use the _identity_ (x + y)⁴ = (x² − y²)² + 4xy(x + y)² which is easily derived by rewriting (x + y)⁴ as (x + y)²(x + y)² = ((x − y)² + 4xy)(x + y)². Letting x + y = s and substituting x² − y² = 24 and xy = 35 into this identity we immediately get s⁴ = 24² + 140s² or s⁴ − 140s² − 24² = 0 Noting that 24 = 2·12 and therefore 24² = 2²·12² and that 12² − 2² = 144 − 4 = 140 we can easily see that this biquadratic equation in s factors as (s² − 12²)(s² + 2²) = 0 so we have s = 12 ⋁ s= −12 ⋁ s = 2i ⋁ s = −2i and the problem is solved. Thus, there are four possible values for x + y = s, two real and two complex.
An efficient way to solve the system x² − y² = 119 xy = 60 for x + y is to use the identity (x + y)⁴ = (x² − y²)² + 4xy(x + y)² which is easily derived by rewriting (x + y)⁴ as (x + y)²(x + y)² = ((x − y)² + 4xy)(x + y)². Letting x + y = s and substituting x² − y² = 9 and xy = 20 into this identity we immediately get s⁴ = 81 + 80s² or s⁴ − 80s² − 81 = 0 It is easy to see that this biquadratic equation in s factors as (s² − 81)(s² + 1) = 0 so we have s = 9 ⋁ s= −9 ⋁ s = i ⋁ s = −i and the problem is solved. Thus, there are four possible values for x + y = s, two real and two complex.
0:42 this is blatantly WRONG! You "fixed it" in your next line but second set of parantheses in this line is wrong.
Hm... so p = -4 (minus four)?
The 3rd line -(p^3 - 4^3) is wrong.
Errore al secondo 47. - (p^3 + 4^3)
I used the rational roots theorem, descartes rule of signs, remainder value theorem, factor theorem, and the upper and lower bounds theorem to solve this. But your solution is so clever
Thankyou
The end with 16+69=80 is somewhat unsatisfying.
the handwriting of "4" looks like "9".
When I compare with the other 4s I can't see that. It seems more likely that he just miswrote. Or maybe in handwriting the first what appears to be a 6 is a 1 😜
Error en el segundo 47....la factorización los signos de los elementos al cubo son incorrectos... El resultado de su agrupamiento daría un 4 al cubo positivo
You would not have the time to solve it in that way in a math competition. Try solving by inspection instead. The first given, x^2 - y^2 = 9, implies (x, y) = (±5, ±4) since (3, 4, 5) is a well known Pythagorean triple. The second given, xy = 20, tells us x and y have the same sign. Therefore (x, y) = (5, 4) or (-5, -4), and correspondingly x + y = 9 or -9.
If you are considering both the positive and negative roots of 9, you should do the same with i, considering ±i as two answers
Wtf x was in front of you both the times. Why need to use these hard equations.
Are kidding?
Ghjjkl v bna v v GC h kuch gy nhi fvnj bnk cc of some people and time in front and address
x × y = 5
16 - (-64) = 80
Right
16+64=80
Nice