X^4/3-12x^2/3+35=0

Question

Find all real solutions of the equation

Answer 1

Let y=x^⅔, then y²=x^4/3.

So the equation becomes y²-12y+35=0=(y-7)(y-5).

Therefore y=7 or 5 and x^⅔=7 or 5.

x²=7³=343 or 5³=125.

x=±√343=±7√7=±18.5203 (approx) or ±√125=±5√5=±11.1803 (approx).

It's probably safer to use the positive versions of these numbers as the real solution for x.