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To solve the equation  in the set of real numbers sqrt(x -1) + sqrt(3 - x) = 2.

I'm not sure, but your request seems a little vague. Or then again it could be my math recall. Based upon what I know, here is how I would solve the problem with the information you have provided me. It's difficult to write calculations out by hand, but I think I've go all the steps in the process listed. Hopefully, this is the information you are searchin for.

The first thing you would need to do is multiply sqrt X (x-1) +  sqrt X (3-x). So you have: sqrtx-sqrt + 3sqrt-sqrtx=2. Then group the terms so you can continue factoring the equation such as: sqrtx-sqrtx-sqrt+3sqrt=2. This would then read:0+2sqrt=2. Divide both sides by 2 such as 2sqrt/2=2/2. You end with sqrt=1.

 

To solve sqrt(x-1)+sqrt(3-x)=2.

Solution: Let us squre both sides of the equation:

x-1+2sqrt[(x-1)(3-x)]+3-x = 4 or

2sqrt[(x-1)(3-x) = 2  . Divide by 2 both sides and then square both sides:

(x-1)(3-x)=1 or

3x-x^2-3+x =1 . Subtract 1 from both sides and then multiply by -1 to make coefficient of x^2  a positive .

x^2-4x+4 = 0.

(x-2)^2 = 0 or

x = 2 is the solution.

First of all, beside of starting to solve the equation, because the equation contains square roots, we have to impose the followings:

x-1>=0, x>=1, x is in [1, inf)

3-x>=0, x=<3, x is in (-inf, 3]

From both, the common interval is [1,3]

In order to ease the solving, we'll move sqrt (3-x) at the right side of the equality, so that:

sqrt(x -1)=2-sqrt(3-x)

After that, we'll square the expression:

x-1=4+3-x-4*sqrt(3-x)

8-2x=4*sqrt(3-x)

2(4-x)=4*sqrt(3-x)

4-x=2*sqrt(3-x)

16-8x+x^2=12-4x

16-8x+x^2-12+4x=0

x^2-4x+4=0

(x-2)^2=0,

x1=x2=2 which belong to [1,3]