x + y² = 1 x² + y³ = 1 from observation, x = 1; y = 0 is a valid solution. considering y ≠ 0: y² = 1 - x y³ = 1 - x² = (1 - x)(1 + x) y = 1 + x y = 1 + x y² = 1 - x (1 + x)² = 1 - x 1 + 2x + x² = 1 - x x² + 3x = 0 x = 0 or x = - 3 if x = 0; y = 1 if x = - 3; y = - 2. so (x, y) ∈ {(-3, -2), (0,1), (1,0)}
Ecuador must not have much internal completion for math Olympiads, because a problem as easy as this wouldn’t even be allowed to be the first question in the non-invitational Brazilian MO
So x= 1-y² Sqaure both sides x²= 1+y⁴-2y² Substituting x² 1+y⁴-2y² + y³ = 1 y⁴+y³-2y²=0 y²(y²+y-2) = 0 y= 0 then x= 1 y= -2 then x= -3 So set of values of xand y (1,0) (0,1) (-3,-2)
The 4th order equation in y strictly follows from the given system of equation. Thus all solutions to the original system also have to satisfy the resulting 4th order polynomial. We calculated all solutions of this equation, so there can be no more solutions for the original system of equations. However, when squaring x in order to substitute into the second equation, this can introduce spurious solutions. From x=1-y^2 it follows that x^2=(1-y^2)^2, but the reverse conclusion is not true. So not all solutions of the 4th order polynomial are guaranteed to be solutions of the original system. We need to verify them all in order to be sure.
I have a question, I had to reach you through this video because it was your latest, how come whenever you have a fraction in the exponent like when you had x^1/x you take the natural log of it, ive never understood the purpose of that, im sure there is one.
Logarithms have a really cool property and that is reducing the order of operations by one. So for example, if you have a product after taking a logarithm you will get a sum. That way it is easier to solve a problem since you can manipulate a product more easily. I hope that helps.
0:10 a nother task like this lets see. 1:32 ok now i tryout to do it by my self und look it afterwords so i will be back to the video within a few seconds after checking nothing xD 1:56 but wait are we talking about real nubers? in this case it works with any x if 0 > y > 0.25 and solution is any real number n for 0.5 > x > 1 if i calculated right but as it looks to easy it is surely false! 3:11 ok i startet with takeing both in a equantion devideing by y and substrackt x to get a quadratic but lets see your way. 7:06 ok so i had a mistake could you maby tell me what was my mistake? 7:09 yes well 7:18 thats funny last time i told sth. simular sombody: beeing evil to an evil person make you to an evil person too. LG K.Furry
Thanks for confirming that guessing is always a good starting point. But usually, guessing only works for the "easy" problems.
Always start with Guess and Check, good job.🤩
Not often I get the right answer before watching the video. Living and learning!
You are a good mathematician .Keep it up !
I appreciate those problems so much. Thanks sir
Your method and solution are so interesting😊, wish you the best
Can't believe I actually found all solutions by inspection
Very nice explanation Sir. Thanks 🙏
Thank you 🦋
After solving the question, I came up with the answer x = -3 and y = -2
Can you please make a video on pentation? That would be great to see.
x + y² = 1
x² + y³ = 1
from observation, x = 1; y = 0 is a valid solution.
considering y ≠ 0:
y² = 1 - x
y³ = 1 - x² = (1 - x)(1 + x)
y = 1 + x
y = 1 + x
y² = 1 - x
(1 + x)² = 1 - x
1 + 2x + x² = 1 - x
x² + 3x = 0
x = 0 or x = - 3
if x = 0; y = 1
if x = - 3; y = - 2.
so (x, y) ∈ {(-3, -2), (0,1), (1,0)}
The sub t = x-1 leads to y^3 = t×(2-t). Division of both eqations gives the quadratic t^2 - 5t +4 = 0. Then the result follows.
Ecuador must not have much internal completion for math Olympiads, because a problem as easy as this wouldn’t even be allowed to be the first question in the non-invitational Brazilian MO
Genial
So x= 1-y²
Sqaure both sides
x²= 1+y⁴-2y²
Substituting x²
1+y⁴-2y² + y³ = 1
y⁴+y³-2y²=0
y²(y²+y-2) = 0
y= 0 then x= 1
y= -2 then x= -3
So set of values of xand y
(1,0) (0,1) (-3,-2)
wait wouldnt it be better to divide the equations?
I have a small question, how can I confirm that there are only three solutions?
The 4th order equation in y strictly follows from the given system of equation. Thus all solutions to the original system also have to satisfy the resulting 4th order polynomial. We calculated all solutions of this equation, so there can be no more solutions for the original system of equations.
However, when squaring x in order to substitute into the second equation, this can introduce spurious solutions. From x=1-y^2 it follows that x^2=(1-y^2)^2, but the reverse conclusion is not true.
So not all solutions of the 4th order polynomial are guaranteed to be solutions of the original system. We need to verify them all in order to be sure.
I see
thx~
I have a question, I had to reach you through this video because it was your latest, how come whenever you have a fraction in the exponent like when you had x^1/x you take the natural log of it, ive never understood the purpose of that, im sure there is one.
Logarithms have a really cool property and that is reducing the order of operations by one. So for example, if you have a product after taking a logarithm you will get a sum. That way it is easier to solve a problem since you can manipulate a product more easily. I hope that helps.
@@frax5051 thank you so much that does help
Hello sir ! I ❤ your videos and teaching can u solve this cubic polynomial equation for me in any video x^(3) - (1/4)x^(2)+0x-1=0
Watch the recent videos on cubic polynomials.
y power of 3 is telling me it should be one more solution
y=0 is a double solution to the original solution. Solutions for y are 0,1, and -2.
x = 1 - y² => x² = y⁴ - 2y² + 1
x² = 1 - y³
y⁴ - 2y² + 1 = 1 - y³
y⁴ + y³ - 2y² = 0
y²(y² + y - 2) = 0
*y = 0 => x = 1*
y² + y - 2 = 0
(y + 2)(y - 1) = 0
*y = - 2 => x = -3*
*y = 1 => x = 0*
Besides trivial, 9−8=1 is the one.
P.S. x²+y³=1 is quite elliptic.
0:10 a nother task like this lets see.
1:32 ok now i tryout to do it by my self und look it afterwords so i will be back to the video within a few seconds after checking nothing xD
1:56 but wait are we talking about real nubers?
in this case it works with any x if 0 > y > 0.25 and solution is any real number n for 0.5 > x > 1 if i calculated right but as it looks to easy it is surely false!
3:11 ok i startet with takeing both in a equantion devideing by y and substrackt x to get a quadratic but lets see your way.
7:06 ok so i had a mistake could you maby tell me what was my mistake?
7:09 yes well
7:18 thats funny last time i told sth. simular sombody: beeing evil to an evil person make you to an evil person too.
LG K.Furry
OMG, what a stupid problem. Let's have this equal 5. Now work it out.