Good solving Sir But from equation (2) you can look for the value of X, i.e from 3⁰=5^½-x, →x=1.236 when I substitute x=1.236 in the initial eqn it gives me 5.415...≈5
Yes you are correct but that solution is rejected because it is above 5. I will do another video on this question due to the comment I have received for clarity sake, just give me some moment sir. Thanks for watching, the observation and commenting sir. Much love @Wirtumromzyreigns....💖💖❤️❤️💕💕
Sres. OnlineMath TV reciban un cordial saludo, gracias este interesante ejercicio, al ser una ecuación de segundo grado, sería dos raíces x1 = 1.17238462 y x2 = -2.21639312. éxitos.
But isn't that the same as saying, for instance, that 12 x 1 = 4 x 3 and then saying that 12 = 4 and 1 = 3? This is obvioulsy not true. I am confused then...
Could you please give a formal proof that 3^X =(5^1/2+X) and 3^0=(5^1/2-X) as pertains to your example? If a times b = c times d, it doesn't necessarily follow that a=c.
Hi. Thank you for your very interesting videos. I have one question. Since there is an x^2 in the equation, should there be 2 solutions? When I used Mathway to solve this equation, it came up with 2 values for x. x ≈ −2.21639311, and 1.17238462. Also, @luisgoncalves raised a question that I find myself asking. Thanks very much.
You can add 3^0 to the LHS, as long as you also add 3^0 to the RHS of the equation. If 3^0 = sqrt(5) - X, then you can add sqrt(5) - X to the RHS. This is assuming 3^0 = sqrt(5) - X... My question is how the heck do you go from 3^X . 3^0 = (sqrt5 + X)(sqrt5 - X) to saying that 3^0 = (sqrt5-X) ? Why would this be true? It can be true, but most often it isn't... I am guessing that's why the end result is not correct @ 1.13... rather than 1.17..., but hey, could be wrong.
In the title, you mention 'Diophantine equation' three times! Wikipedia: "a Diophantine equation is an equation, ... such that the only solutions of interest are the integer ones" I doubt your solution is integer...
Unfortunately, this time it doesn't work with a simple equation from the product of the quadratic equation. The result obtained in this way is somewhat too small. Try again with the approach x1 = x0 - f(x0)/f'(x0). The numerical result is approximately 1.17238462
Of a truth i got 1.17238.... and -2.2163....using the graphical method. I will resolve this and if possible take this video down. Thanks for the comment sir.
@@onlineMathsTV I hope you don't take this down even if it turns out incorrect. There are many interesting techniques you show here. What I would like to see is you doing another video explaining why this doesn't work. Where exactly it went wrong (assuming it's wrong). There's an expression I remember from my school days -- "Even Homer nods." Even the most highly skilled can make an occasional mistake, and then learn from it -- and show us how to learn from our mistakes. Thanks again for your great videos.
I love your way of solving such problems. Good luck. This is really genious.
Compliments!
Excellent
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thank you very much, it's beautiful
I'm glad you like it and thank you for watching and at the same time commenting sir.
We all @Onlinemathstv love you dearly 💕💕💕💖💖💖❤️❤️
Good solving Sir
But from equation (2) you can look for the value of X, i.e from 3⁰=5^½-x, →x=1.236
when I substitute x=1.236 in the initial eqn it gives me 5.415...≈5
Yes you are correct but that solution is rejected because it is above 5.
I will do another video on this question due to the comment I have received for clarity sake, just give me some moment sir.
Thanks for watching, the observation and commenting sir.
Much love @Wirtumromzyreigns....💖💖❤️❤️💕💕
👍👍👍
Sres. OnlineMath TV reciban un cordial saludo, gracias este interesante ejercicio, al ser una ecuación de segundo grado, sería dos raíces x1 = 1.17238462 y x2 = -2.21639312. éxitos.
💯
Thanks a million sir.
But isn't that the same as saying, for instance, that 12 x 1 = 4 x 3 and then saying that 12 = 4 and 1 = 3? This is obvioulsy not true. I am confused then...
Nope, you cannot equate constants in this regard
@@onlineMathsTV Hi there, but that's what you just did, isn't it?
3^1.1328+1.1328^2=4.754 (凸)
Yes, but that's not a Diophantine Equation.
By definition, a Diophantine Equations is when solutions are integers.
noted
Your equation 2 is 3^0=5^(1/2)- x
So we can use this equation only to solve for x.
X=5^(1÷2)-3^0=1.2361
Yes, because the second value of x from the second equation is rejected
Could you please give a formal proof that 3^X =(5^1/2+X) and 3^0=(5^1/2-X) as pertains to your example? If a times b = c times d, it
doesn't necessarily follow that a=c.
We’ll be has no proof because this statement is just false just plug in 1
Hi. Thank you for your very interesting videos. I have one question. Since there is an x^2 in the equation, should there be 2 solutions? When I used Mathway to solve this equation, it came up with 2 values for x. x ≈ −2.21639311, and 1.17238462. Also, @luisgoncalves raised a question that I find myself asking. Thanks very much.
Por graficas de la exponencial y la cuadratica hay dos soluciones reales
Clever
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Much love from Onlinemathstv 💕💕💕❤️❤️❤️
X=sq rt of 5 minus 1
The caliculator isn't allowed in the EXAM. Why did u?
2 values of x
Eeesi
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I lost you at 5:30 when you added both equations (i) and (ii) as if they were independent
You can add 3^0 to the LHS, as long as you also add 3^0 to the RHS of the equation. If 3^0 = sqrt(5) - X, then you can add sqrt(5) - X to the RHS. This is assuming 3^0 = sqrt(5) - X...
My question is how the heck do you go from 3^X . 3^0 = (sqrt5 + X)(sqrt5 - X) to saying that 3^0 = (sqrt5-X) ? Why would this be true? It can be true, but most often it isn't... I am guessing that's why the end result is not correct @ 1.13... rather than 1.17..., but hey, could be wrong.
Another video on this will be produce for clarification sir.
Actually I got two different answers graphically.
I will resolve this.
Thanks for your comment sir.
youlosex=-2.2164
You are lost X=-2.2164
Some of the assumptions not clear to me
I will reproduce this video clip pls.
Thanks
In the title, you mention 'Diophantine equation' three times!
Wikipedia: "a Diophantine equation is an equation, ... such that the only solutions of interest are the integer ones"
I doubt your solution is integer...
Unfortunately, this time it doesn't work with a simple equation from the product of the quadratic equation. The result obtained in this way is somewhat too small. Try again with the approach x1 = x0 - f(x0)/f'(x0). The numerical result is approximately 1.17238462
Of a truth i got 1.17238.... and -2.2163....using the graphical method.
I will resolve this and if possible take this video down.
Thanks for the comment sir.
@@onlineMathsTV I hope you don't take this down even if it turns out incorrect. There are many interesting techniques you show here. What I would like to see is you doing another video explaining why this doesn't work. Where exactly it went wrong (assuming it's wrong). There's an expression I remember from my school days -- "Even Homer nods." Even the most highly skilled can make an occasional mistake, and then learn from it -- and show us how to learn from our mistakes. Thanks again for your great videos.
Sorry, this solution is incorrect.
excuse me you are wrong the true solutions are : x1 = -2.21 , x2= 1.17 . every one can test
ok
😢It‘s false!
Ok sir, I will resolve this again.
Thanks for the observation sir.