3x312 x512 find xSolutionWell substitute the exponent 12 by

(3x+3)^1/2 = (x+5)^1/2    find x

Solution

We\'ll substitute the exponent 1/2 by sqrt and we\'ll re-write the equation:

sqrt(3x+3)= sqrt(x+5)

Now, we\'ll impose the conditions of existence of the square roots:

3x + 3>=0

We\'ll divide by 3:

x + 1>=0

x>=-1

x + 5>=0

x>=-5

The common interval of admissible values is [-1,+inf)

If we\'ll square raise, we\'ll eliminate the square roots:

[sqrt(3x+3)]^2 = [sqrt(x+5)]^2

3x + 3 = x + 5

We\'ll isolate x to the left side. For this reason, we\'ll subtract x and 3 both sides:

3x - x = 5 - 3

2x = 2

We\'ll divide by 2 and we\'ll get:

x = 1

Since 1 belongs to the interval of admissible values, the solution x = 1 is valid.