# sqr(x+5-4sqr(x+1))+sqr(x+2-2sqr(x+1))=1x=?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll subtract `sqrt(x+2-2sqrt(x+1))`  both sides:

`sqrt(x+5-4sqrt(x+1)) = 1 - sqrt(x+2-2sqrt(x+1))`

We'll raise to square both sides to remove the square root:

`x + 5 - 4sqrt(x+1) = 1 - 2sqrt(x + 2-2sqrt(x+1)) + x + 2 - 2sqrt(x+1)`

We'll isolate the term `2sqrt(x + 2-2sqrt(x+1))`  to the right side:

`x + 5 - 4sqrt(x+1) - 1 - x - 2 + 2sqrt(x+1) = - 2sqrt(x + 2-2sqrt(x+1))`

We'll eliminate like terms:

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

We'll divide both sides by -2:

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

We'll raise again to square to eliminate the radical from the right side:

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

`x + 2 - 2sqrt(x+1) - x - 2 + 2sqrt(x+1)) = 0`

We'll eliminate all terms and we'll get:

0 = 0

Therefore, any real value of x, comprised within interval `[-1,+oo)`  is a solution of the equation,

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