# (629-x)^1/4+(77+x)^1/4=8 x=?

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### 1 Answer

Solve `root(4)(629-x)+root(4)(77+x)=8` .

I believe that this problem was posed as a guess and check problem. We see that 629 is 4 away from a perfect fourth power, so we try x=4.

`root(4)(625)+root(4)(81)=5+3=8` **so x=4 is a solution.**

Trying to get 3+5=8 we find **x=548 is a solution** since `root(4)(81)+root(4)(625)=3+5=8`

An appeal to the graph convinces that these are the only solutions:

** Solving by algebra requires repeated squaring:

`root(4)(629-x)+root(4)(77+x)=8`

`sqrt(629-x)+2root(4)(629-x)root(4)(77+x)+sqrt(77+x)=64`

`sqrt(629-x)+sqrt(77+x)=64-2root(4)(629-x)root(4)(77+x)`

`2sqrt(629-x)sqrt(77+x)+706=4sqrt(629-x)sqrt(77+x)-256root(4)(629-x)root(4)77+x)+4096`

`-x^2+552x-12994sqrt(629-x)sqrt(77+x)=-2921458`

`-x^2+552x+2921458=12994sqrt(629-x)sqrt(77+x)`

`x^4-1104x^3+163305824x^2-89976618240x+357293650176=0 `

**The real solutions to this are x=4 and 548 as above.**