Real values of x in equationDetermine all real values of x in the equation square root (x-1) =1 – square root(2-x)?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to solve the equation: sqrt (x-1) =1 – sqrt(2-x)

sqrt (x-1) =1 – sqrt(2-x)

square both the sides

x - 1 = 1 + 2 - x - 2*sqrt (2 - x)

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

=> x - 2 = -1*sqrt(2 - x)

square both the sides

x^2 + 4 - 4x = 2 - x

=> x^2 - 3x + 2 = 0

=> x^2 - 2x - x + 2 = 0

=> x(x - 2) - 1(x - 2) = 0

=> (x - 1)(x - 2) = 0

=> x = 1 and x = 2

The required real solutions are x = 1 and x = 2.

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 real solutions of the equation.

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