Tried the Hardest Algebra Olympiad Question | You Should Try Too!
Vložit
- čas přidán 13. 09. 2024
- Tried the Hardest Algebra Olympiad Question | You Should Try Too!
Welcome to infyGyan! In this video, we present an incredible radical equation that will test our algebra skills and challenge our mathematical thinking. This problem is perfect for those preparing for Math Olympiad or anyone who loves tackling hard math problems.
Join us as we break down the steps to solve this intriguing equation, and see if you can find the solution on your own. Don't forget to leave your answers and approach in the comments below!
If you enjoy solving complex math problems, make sure to subscribe to our channel and hit the notification bell for more exciting math challenges.
Additional Resources:
• A Tricky Radical Equat...
• A Nice Radical Equatio...
• Can You Solve This Rad...
• A Nice Factorial Equat...
🔔 Don't forget to subscribe and hit the bell icon to stay updated with our latest math challenges and tutorials. If you enjoy the video, please give it a thumbs up and share it with your friends!
💬 Comment below if you have any questions or if you were able to solve the problem. We love hearing from our viewers!
Thank you for watching and happy problem-solving!
that was a tough one. Great problem and solution.
Let [13-x^2]!/2= [19-x^3]^1/3 = y. Then, x^2+y^2 = 13 and x^3+y^3=19. Let x=y=a and xy=b. Then, a^2-2b=13 > b = 1/2(a^2-13). Again, a[a^2-3b]=19.> a[a^2-3/2(a^2-13)] =19 > a^3-39a+38=0 > a=1 > b = -6. Thus, x-6/x=1 > x^2-x-6 =0 > x=-2,3. But x=3 is not a valid solution. So, x=-2.
let a = (13 - x^2)^1/2 , b = (19 - x^3)^1/3 (a,b > 0) then a = b , a^2 = 13 - x^2 , b^3 = 19 - x^3
=> a^2 + x^2 = (a + x)^2 - 2ax = 13 , a^3 + x^3 = (a + x)^3 - 3ax(a + x) = 19
let p = a + x, q = ax
=> p^2 - 2q =13 ---(1) p^3 - 3pq = 19 ---(2)
p*(1) - (2) => pq = 13p - 19, putting this to (2)
=> p^3 - 3(13p - 19) = 19 => p^3 - 39p + 38 = (p - 1)(p^2 + p - 38) = 0
(case p = 1) => q = (p^2 - 13)/2 = -6, a,x are roots of t^2 - t - 6 = (t -3)(t + 2) = 0
since a > 0, (a,x) = (3, -2) => x = -2
(case p^2 + p - 38 = 0) a,x are roots of t^2 -pt + q =0
determinant(D) = p^2 - 4q = p^2 - 4*(p^2 - 13)/2 = -p^2 + 26 p = (-1 ± 3√17)/2, p^2 = 38 - p
D = -p^2 + 26 = (p -38) + 26 = p -12 = (-1 ± 3√17)/2 -12 = (-25 ± 3√17)/2 < 0
no real solution in this case.
Let y=sqrt(13-x^2), and we change these two equations into:
x^2+y^2=13
x^3+y^3=19
Let s=x+y and p=xy, and we have:
s^2-2p=13
s^3-3ps=19
Plug p=(s^2-13)/2 into the second equation, and we have:
s^3-39s+38=0
This can be solved easily.
s_1=1, s_2=(-1+sqrt(152))/2, s_3=(-1-sqrt(152))/2.
Note that s=x+y and its max is sqrt(2)*sqrt(13) and thus s_2 and s_3 are rejected.
Therefore s=1 and p=-6. That is:
x+y=1
xy=-6.
And y>=0.
This gives us x=-2.
Θετω 19-χ^3=y^3 και εχω (13-χ^2)^(1/2)=y 13-χ^2=y^2 και εχω το συστημα χ^3+ y^3=19 και χ^2+y^2=13 απο τη λυση του συστηματος και θετοντας χ+ y=α και χy=β στο συνολο Ζ εχω λυση χ=-2. (Η λυση χ=3 απορριπτεται).
The left member is a semicircle, and the right member is decreasing. Looking at the x and y intercepts, you will see only one solution exists. That solution must be negative, between -sqrt(13) and zero. This is where I'd look at integers: -3, -2, -1. Yep, x=-2 is the only real solution.
Only one real solution x=-2
X=-2 ; ....
Other solution
(13)^2 ➖, (x^4)^2={169 ➖ x^4}= 165 10^10^65^1 2^52^5^1^1 1^12^1 2^1 (x ➖ 2x+1) (x^3 )^2➖ (19)^2 ➖ (x^3)^2={x^9 ➖ 361} ➖ x^9={352 ➖ x^9}= 343 10^30^43^1 10^30^1^1 10^5^6 2^5^5^3^2 1^1^13^2 (x ➖ 3x+2).
The given equation equivalent with
[(13 - x^2)^1)2]^6 = [(19 -x^3)^1/3]^6
(13 - x^2)^3 = (19 - x^3)^2
2197 - 507 x^2 + 39 x^4 - x^6 = 361 - 38 x^3 + x^6 or
2 x^6 - 39 x^4 - 38 x^3 + 507 x^2 - 1836 = 0 or
( x +2 )( x - 3)(2 x^4 + 2 x^3 - 25 x^2 - 51 x + 306) = 0.
So x = - 2 and x = 3 . .
Horrible method that puts you in a bad situation to solve Hexic polynomial. You have not shown that the remaining quartic has 4 imaginary solutions so that you can disregard it. Also only x=-2 works. You have obtained an extraneous solution with the other one.