Question

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

Answer 1

Answer:

x = 1 and x = 5

Step-by-step explanation:

$\sqrt{3x+1}$ - $\sqrt{x-1}$ = 2

First, because there are two square roots, we have to determine domain of function y = $\sqrt{3x+1}$ - $\sqrt{x-1}$ .

3x + 1 ≥ 0 ==> x ≥ - $\frac{1}{3}$

x - 1 ≥ 0 ==> x ≥ 1

So, in set of real number the function exist if x ≥ 1 .

($\sqrt{3x+1}$ - $\sqrt{x-1}$)² = 2²

($\sqrt{3x+1}$)² - 2$\sqrt{3x+1}$ × $\sqrt{x-1}$ + ($\sqrt{x-1}$)² = 4

3x + 1 - 2$\sqrt{(3x+1)(x-1)}$ + x - 1 = 4

4x - 2$\sqrt{(3x+1)(x-1)}$ = 4

- 2$\sqrt{(3x+1)(x-1)}$ = 4 - 4x

$\sqrt{(3x+1)(x-1)}$ = 2x - 2

($\sqrt{(3x+1)(x-1)}$)² = (2x - 2)²

(3x + 1)(x - 1) = 4x² - 8x + 4

3x² - 2x - 1 - 4x² + 8x - 4 = 0

- x² + 6x - 5 = 0

x² - 6x + 5 = 0

D = 36 - 20 = 16

$x_{12}$ = (6 ± √D) / 2 = 3 ± 2

$x_{1}$ = 1

$x_{2}$ = 5

Answer 2

x = 1 or 5

Step-by-step explanation:

√(3x+1) - √(x-1) = 2

Square on both sides, we get:

[√(3x+1) - √(x-1)]²  = 2²

  (3x+1) + (x-1) -2√(3x+1)(x-1) = 4

 4x -2√(3x+1)(x-1) = 4

  4x - 4 =  2√(3x+1)(x-1)

  2x -2 = √(3x+1)(x-1)

   (2x -2)² = [√(3x+1)(x-1)]²

   4x² + 4 - 8x = (3x+1)(x-1)

    4x² + 4 - 8x = (3x² +x - 3x - 1)

    x² - 6x + 5 = 0

    x² - 5x -x + 5 = 0

    x(x-5) -1(x-5) = 0

   (x-5) (x-1) = 0

  Therefore x = 1 or 5.