Determine all real values of x in the equation (x-1)^(1/2)+(2-x)^(1/2) =1 ?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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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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neela | High School Teacher | (Level 3) Valedictorian

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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.

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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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