A Nice Radical Equation with Fifth Root

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[pwmiles]

Guessing symmetry would be involved I framed the problem as
y = (2 - x)^(1/5)
2 = (y - x)^(1/5)

With y = 2 these become the same equation so you only have to solve
2 = (2 - x)^(1/5)
2 - x = 32
x = 2 - 32
x = -30

Of course, there could be other roots, but since x + 32 is increasing and (2 - x)^(1/5) is decreasing, the graphs cross in only one place so there are no other real roots.

[gdtargetvn]

Note: at the "y-32=2-y^5" part, rewrite it as y^5+y=2^5+2
You may notice the term is f(x)=x^5+x, which is an increasing function, so it's injective
so, f(y)=f(2) => y=2

[samanzakeri]

i have an easier way of solving this. ( this way our answer domain is a little bigger than what we want but we can plug our answers in the equation to check ).
so if a = b then we can say b is divisible by a and also with the same rules we can say b^5 is also divisible by a so we can apply this in the equation and we get this : x + 32 = root 5 of (2 - x) => x + 32 | root 5 of (2-x) ( btw a|b means b is divisable by a ) => x + 32 | 2-x now x + 32 is also divisable by itself and if a|b and a|c then a|b-c and a|b+c and a|c-b now using this we have x + 32|x-2 and x+32|x+32 so we get x+32|34 now what numbers can divide 34? well that would be 1, 2, 17, 34 and -1, -2, -17, -34 which are equal to x+32 so x can be -31, -30, -15, 2, -33, -34, -49, -66 now we plug each of these in x in the equation and see that only -30 satisfies the equation. btw im in tenth grade and i haven't learned complex math yet so idk how to solve for complex answers.

[vaggelissmyrniotis]

If we take f(x)=(x+32)^5+x-2 its obvious root is x=-30 and f'(x)>0 so it will be f(x)>0 or (x+32)^5>2-x for x>-30 and then we take the fifth root on both sides.We do the same when x<-30.

[rickdesper]

The solution at x=-30 is pretty obvious. (I see a fifth root function so I want to put a fifth power underneath. 32=2^5 is the obvious choice.) Once you see one solution you can see that the function on the left is strictly increasing while the one on the right is strictly decreasing, so there will be at most one solution. Save yourself five minutes of algebraic manipulation.
Don't underrate "guess and check".

[neuralwarp]

You could just say "see you next time".

[samppufs]

I just guessed it and then showed the derivative is always negative so the function is always decreasing therefore having only 1 real solution

[scottleung]

I got x=-30 by plugging it in. I knew that by dividing the resulting quintic on the LHS by x+30 after raising both sides to the fifth power, I would get a quartic whose signs are all positive, thus 4 complex solutions making -30 the only real solution.

[Szu]

Is there any simple way to find complex solutions?

[georgerenko]

Your answer x=-30 is wrong. Correct answer is: x=-34