# What is root of equation (x+2-4(x-2)^(1/2))^(1/2)+(x+7-6(x-2)^(1/2))^(1/2)-1=0?

You should come up with the following substitution, such that:

`(x - 2)^(1/2) = y => y^2 = x - 2 => x = y^2 + 2`

Replacing the variable,  in the given equation, yields:

`sqrt(y^2 + 2 + 2 - 4y) + sqrt(y^2 + 2 + 7 - 6y) - 1 = 0`

`sqrt(y^2 - 4y + 4) + sqrt(y^2 - 6y + 9) = 1`

You need to replace `(y - 2)^2 ` for `y^2 - 4y + 4` and ` (y - 3)^2` for `y^2 - 6y + 9` , such that:

`sqrt((y - 2)^2) + sqrt((y - 3)^2) = 1`

`|y - 2| + |y - 3| = 1`

Using the definition of absolute value, yields:

`|y - 2| = {(y - 2, y>=2),(2 - y, y<2):} `

`|y - 3| = {(y - 3, y>=3),(3 - y, y<3):}`

You need to solve the equation for `y in (-oo,2)` , such that:

`2 - y + 3 - y = 1 => -2y = -4 => y = 2 !in (-oo,2)`

You need to solve the equation for `y in [2,3)` , such that:

`y - 2 + 3 - y = 1 => 1 = 1 => y in [2,3)`

You need to solve the equation for `y in [3,+oo)` , such that:

`y - 2 + y - 3 = 1 => 2y = 6 => y = 3 in [3,+oo)`

You need to solve for `x` the following inequality, such that:

`2<=sqrt(x - 2) <= 3 => 4<= x - 2 = 9 => 6 <= x <= 11`

Hence, evaluating the solutions to the given equation yields that the equation holds for all real values `x in [6,11].`

Approved by eNotes Editorial Team

The root of the equation `(x+2-4(x-2)^(1/2))^(1/2)+(x+7-6(x-2)^(1/2))^(1/2)-1=0` has to be determined.

`(x+2-4(x-2)^(1/2))^(1/2)+(x+7-6(x-2)^(1/2))^(1/2)-1=0`

=> `sqrt(x-4*sqrt(x-2)+2) + sqrt(x-6*sqrt(x-2)+7) = 1`

Take the square of both the sides

`2*sqrt(x-6*sqrt(x-2)+7)*sqrt(x-4*sqrt(x-2)+2)` +`2*x-10*sqrt(x-2)+9` = 1

`sqrt(x-6*sqrt(x-2)+7)*sqrt(x-4*sqrt(x-2)+2)` +`x-5*sqrt(x-2)+4` = 0

`x-5*sqrt(x-2)+4` = `-sqrt(x-6*sqrt(x-2)+7)*sqrt(x-4*sqrt(x-2)+2)`

Take the square of both the sides

`x^2+33*x+(-10*x-40)*sqrt(x-2)-34` = `x^2+33*x+(-10*x-40)*sqrt(x-2)-34 `

=> 0 = 0

This equation is true for all values of x

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