Verified answer

Something is wrong here.

If we square both sides of the equation we get: x+3 +x-1=1 which simplifies to 2x + 2 = 1 which converts into 2x = -1 which is x = -0.5

However, when we substitute -0,5 for x in the original equation, we end up trying to find the square root of a negative number.

rt(x+3) + rt(x-1)=1

The trick to solving these things is to isolate the radical term to one side of the

equation, then square both sides. Sometimes you have to do this procedure

twice , as is the case in this problem.

rt(x+3)+rt(x-1)=1

rt(x+3)=1-rt(x-1)

Square both sides (use FOIL on the right hand term)

x+3= [1-rt(x-1)][1-rt(x-1)]

x+3= 1-2rt(x-1)+(x-1)

x+3=1-2rt(x-1)+x-1

x+3=-2rt(x-1)+x

x-x+3= -2rt(x-1)

3= -2rt(x-1)

3/-2=rt(x-1)

Square both sides

9/4=(x-1))

9=4(x-1)

9=4x-4

9+4=4x

13=4x

13/4=x

We have to verify our answer, because in some cases the squaring operation

makes negative values become positive and the resulting answer isn't right.

rt(x+3)+rt(x-1)=1

For x= 13/9,

rt(13/4+12/4)+ rt(13/4-4/4)

=rt(25/4) + rt(9/4)

=5/2+3/2

=8/2

=4

And 4 does not = 1

For this problem, no solution.

square both sides.

x+c + c-1 = 1

x +2c -1 = 1

add 1 to both sides

x + 2c =0

subtract 2c from both sides

x=-2c

go back and substitute x with -2c

(-2c) + 3 + (-2c) -1 = 1

-4c+2=1

subtract 2 from both sides

-4c=-1

divide by -4

c=1/4

put c into x..

x=-2(1/4)

x=-1/2

unsolvable you probably are looking to isolate a variable in any case there is no solution with information given.GL