Olympiad Mathematics | Solving A Radical Equation With Imaginary Number

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[@WorldwideBibleClass-qr9jk]

Cool

[@danielfranca1939]

Nice video beother, wishing you the best on this journey.

[@doganhalli2690]

Put the outermost x into the innermost square root . After that do the same thing for other x ‘s and then multiply them you get eighthrootx^15=7i Take the 8th power of both sides.

[@RealQinnMalloryu4]

(x ➖7 ix+7i )

[@Mathprowess]

How?

[@spaghetti1383]

You did not solve the equation, your solution is extraneous. If you plug 7^(8/15) into the left hand side of the original equation, you get a positive real number, not an imaginary number.
To get the actual solution, you need to know about polar form of complex numbers or roots of unity. The magnitude of the the solution is indeed r=7^(8/15), but the direction is not a=0 (positive real number line). x=r*e^(ia) where (e^(ia))^15/8 = i. Since i = e^(pi*i/2), then 15/8*ia = pi*i/2. Now we can solve for a = 4pi/15. So x = 7^(8/15)*e^(4pi*i/15) = 7^(8/15)*(cos(4pi/15)+isin(4pi/15)). This is just the principle solution, there might be more with different directions.

[@Mathprowess]

You are very right. Thanks for the observation sir.