# Determine root of equation square root(x+2square root(x-1))+root(x-2square root(x-1))=square root 2?

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`sqrt(x+2sqrt(x-1))+sqrt(x-2sqrt(x-1))=sqrt(2)`

Square both sides -- note that the left side is a binomial and will have 4 terms; also squaring might introduce extraneous solutions so we must check any answer to see that it is a solution in the original problem.

`x+2sqrt(x-1)+2sqrt((x+2sqrt(x-1))(x-2sqrt(x-1)))+x-2sqrt(x-1)=2` Add like terms and simplify the radicand:

`2x+2sqrt(x^2-4(x-1))=2`

Divide both sides by 2 and rewrite the radicand as a perfect square:

`x+sqrt((x-2)^2)=1`

`x+|x-2|=1`

`|x-2|=1-x`

If x>2 then x-2=1-x ==> 2x=3 ==> `x=3/2`

If x<2 then -x+2=1-x ==> 2=1 which is impossible.

The only possible solution is `x=3/2` .

Check in the original equation:

`sqrt(3/2+2sqrt(3/2-1))+sqrt(3/2-2sqrt(3/2-1))`

`=sqrt(3/2+2sqrt(1/2))+sqrt(3/2-2sqrt(1/2))`

`=sqrt(3/2+sqrt(2))+sqrt(3/2-sqrt(2))`

=2 not `sqrt(2)`

**So there is no solution.**

The graphs -- note that they do not intersect: