# 1) 4*sqrt(5-2*sqrt(6)) 2)2x+sqrt(3x-2)=3

1) You need to raise to square the product to remove the square root such that:

`4*sqrt(5-2*sqrt(6)) = 16(5 - 2sqrt6)`

You need to open the brackets such that:

`16(5 - 2sqrt6) = 80 - 32sqrt6`

`16(5 - 2sqrt6) = 80 - 32*2.449`

`16(5 - 2sqrt6) ~~ 1.616`

Hence, evaluating...

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1) You need to raise to square the product to remove the square root such that:

`4*sqrt(5-2*sqrt(6)) = 16(5 - 2sqrt6)`

You need to open the brackets such that:

`16(5 - 2sqrt6) = 80 - 32sqrt6`

`16(5 - 2sqrt6) = 80 - 32*2.449`

`16(5 - 2sqrt6) ~~ 1.616`

Hence, evaluating the value of the square root yields 1.616.

2) You need to solve the equation `2x+sqrt(3x-2)=3` , hence you need to keep the square root to the left side and to move 2x to the right such that:

`sqrt(3x-2) = 3 - 2x`

You need to raise to square both sides such that:

`3x - 2 = 9 - 12x +4x^2`

`4x^2 - 12x + 9 - 3x + 2 = 0`

Collecting like terms yields:

`4x^2 - 15x + 11 = 0`

You need to use quadratic formula such that:

`x_(1,2) = (15+-sqrt(225 - 176))/8`

`x_(1,2) = (15+-7)/8 =gt x_1 = 22/8 = 11/4`

`x_2 = 8/8 = 1`

Notice that the expression under radical needs to be positive, hence `3x - 2 gt= 0 =gt x gt= 2/3`

Since both roots of equation are larger than `2/3` , hence, the solutions to the equation `2x+sqrt(3x-2)=3`  are `x = 1`  and `x = 11/4.`

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