Mathematical Olympiad | Learn how to solve radical equation quickly | Math Olympiad Training

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Mathematical Olympiad | Learn how to solve radical equation quickly | Math Olympiad Training

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[عبدالفتاحالسيد-رر]

Thank you very much for this explanation.
Greetings to you from Syria
I wish to visit the United States of America.

[williamwingo]

I did the same substitution but just went straight to the quadratic formula:
let u = x^(1/6); then x^(1/3) = u^2:
3u^2 – 7u – 6 = 0; invoking the quadratic formula with a = 3, b = –7, and c = –6:
u = ((–(–7)) +/– (sqrt(((--7)^2)) – 4*3*(–6)))/(2*3);
= (7 +/– (sqrt(49 – (–72))))/6;
= (7 +/– (sqrt(49 + 72)))/6;
= (7 +/– sqrt(121))/6;
= (7 +/– 11)/6;
= either 18/6 or –4/6; that is, either 3 or –2/3.
Then in the first case, x = u^6 = 3^6 = (3^3)^2 = 27^2 = 729.
[Edited to balance the parentheses.]
Cheers. 🤠

[GuglielmoArmillei]

Can you solve Integral (from 0 to pi/2) of: (x dx) / tanx ?

[kumarsunil]

Sir...with your kind permission I would like to change a statement given by you ...x power 1/6 is not an exponential function it would be even root function and which could also not be negative... please reply..

[ashieshsharmah]

Sir can you please tell me why x^1/6≠ -2/3 . As because -2/3 is not the value of X it's is a value of 6th root of X . Which can be both positive and negative as well . If it was x^6 =-2/3 then is not defined (imaginary) . Please help me out .🙏

[giuseppemalaguti]

x=3^6....poi ci sarebbe una soluzione, x=(-2/3)^6=4(ko)

[HappyFamilyOnline]

Very interesting👍
Thanks for sharing😊