# r=?Solve for r value: 2 = sqrt(3x - 2) - sqrt(10-x)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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3x - 2 >= 0

3x > =2

x>=2/3

10 - x>=0

x=<10

The interval of admissible values for x is [2/3 ; 10].

Now, we'll solve the equation. We'll multiply both sides by the conjugate of the expression sqrt(3x - 2) - sqrt(10-x).

2 = sqrt(3x - 2) - sqrt(10-x) (1)

2[sqrt(3x- 2) + sqrt(10-x)] = 3x - 2 - 10 + x

We'll combine like terms:

2[sqrt(3x - 2) + sqrt(10-x)] = 4x - 12

We'll divide by 2:

[sqrt(3x - 2) + sqrt(10-x)] = 2r - 6 (2)

sqrt(3x - 2) - sqrt(10-x) + sqrt(3x - 2) + sqrt(10-x) = 2 + 2x - 6

We'll combine and eliminate like terms:

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

We'll divide by 2:

sqrt(3x - 2) =x - 2

We'll square raise both sides:

3x - 2 = (x-2)^2

3x - 2 = x^2 - 4x + 4

We'll use the symmetric property and we'll combine like terms:

x^2 - 4x + 4 - 3x + 2 = 0

x^2 - 7x + 6 = 0

x1 = 1

x2 = 6

Since both solutions belong to the interval of admissible values, they are accepted.