Determine all real values of x in the equation (x-1)^(1/2)+(2-x)^(1/2) =1 ?
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We have to solve the equation : (x-1)^(1/2)+(2-x)^(1/2) =1
(x-1)^(1/2) + (2-x)^(1/2) = 1
square both the sides
=> [(x-1)^(1/2) + (2-x)^(1/2)]^2 = 1^2
=> (x - 1) + (2 - x ) + 2[(x-1)(x-2)]^(1/2) = 1
=> 1 + 2[(x-1)(x-2)]^(1/2) = 1
=> 2[(x-1)(x-2)]^(1/2) = 0
=> (x - 1) (x - 2) = 0
x is equal to 1 and 2.
Therefore the required values of x are 1 and 2.
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To determine all real values of x in the equation
(x-1)^(1/2)+(2-x)^(1/2) =1 .
(x-1)^(1/2) = 1-(2-x)^(1/2)
We square both sides:
x-1 = 1- 2(2-x)^(1/2) +(2-x).
x-1-1- 2+x = -2(2-x)^1/2).
2(x-2) = -2(2-x)^(1/2).
We divide by 2:
(x-2) = -(2-x)^(1/2).
We square both sides:
(x-2)^2 = (2-x).
(x-2)^2 +(x-2) = 0.
(x-2) (x-2+1) = 0
(x-2) (x-1) = 0.
x-2 = 0, or x-1 = 0.
x = 2 , or x = 1.
We'll impose the constraints of existence of square roots:
x - 1>=0
x >= 1
2 -x >=0
x =<2
The interval of admissible values for x is [1 ; 2].
Now, we'll solve the equation. We'll subtract sqrt(2-x) both sides:
sqrt(x-1) = 1 - sqrt(2-x)
We'll raise to square both sides:
x - 1 = 1 - 2sqrt(2-x)+ 2 - x
We'll combine like terms:
x - 1 = 3 -x - 2sqrt(2-x)
We'll subtract 3 - x both sides:
2x - 4 = -2sqrt(2-x)
We'll divide by -2:
2 - x = sqrt(2-x)
We'll raise to square both sides:
4 - 4x + x^2 = 2 - x
We'll subtract 2 - x:
x^2 - 3x + 2 = 0
(x-1)(x-2) = 0
x-1=0
x = 1
x-2=0
x = 2
Since both values belong to the interval of admissible values, we'll accept them as solutions of the equation.
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