Question

under root 3x+1 under root x-1 =2​

Answer 1

Step-by-step explanation:

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

3x + 1 ≥ 0 ==> x ≥ - \frac{1}{3}31

x - 1 ≥ 0 ==> x ≥ 1

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

(\sqrt{3x+1}3x+1 - \sqrt{x-1}x−1 )² = 2²

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

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

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

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

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

(\sqrt{(3x+1)(x-1)}(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}x12 = (6 ± √D) / 2 = 3 ± 2

x_{1}x1 = 1

x_{2}x2 = 5