# (8+x)^(1\4)+ (89-x)^(1\4)= 5 x=?

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Solve `(8+x)^(1/4)+(89-x)^(1/4)=5` or `root(4)(8+x)+root(4)(89-x)=5` .

This problem had to have been posed as a guess and check problem. We know that 2+3=5; 2 is the fourth root of 16 and 3 is the fourth root of 81.

**We notice that 8+8=16 and 89-8=81 so x=8 is a solution.**

**We notice that 8+73=81 and 89-73=16 so x=73 is a solution.**

We can appeal to a graph to show that these are the only solutions:

** To solve using algebra requires multiple squaring, and then solving a quartic:

`root(4)(8+x)+root(4)(89-x)=5`

`sqrt(8+x)+2root(4)(8+x)root(4)(89-x)+sqrt(89-x)=25`

`sqrt(8+x)+sqrt(89-x)=25-2root(4)(8+x)root(4)(89-x)`

`8+x+2sqrt(8+x)sqrt(89-x)+89-x=625-100root(4)(8+x)root(4)(89-x)+4sqrt(8+x)sqrt(89-x)`

`-528=-100root(4)(8+x)root(4)(89-x)+2sqrt(8+x)sqrt(89-x)`

`sqrt(8+x)sqrt(89-x)-264=-50root(4)(8+x)root(4)(89-x)`

`-x^2+81x+528sqrt(8+x)sqrt(89-x)+70408=2500sqrt(8+x)sqrt(89-x)`

`-x^2+81x+70408=1972sqrt(8+x)sqrt(89-x)`

`x^4-162x^3+3754529x^2-303585408x+2188472256=0`

**The two real solutions are 8 and 73 as above.**