# (3x+3)^1/2 = (x+5)^1/2 find x

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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.**

(3x+3)^1/2 = (x+5)^1/2

first in order for the equality to be valid , 3x+3 should be greater that0 and x+5 >= 0

==> 3x + 3 >= 0 and x+ 5 >=0

==> x >= -1 and x >=-5

==> x >= -1

Let us square both sides:

==> (3x+3)^1/2]^2 = [(x+5)^1/2]^2

==> (3x+3) = x+5

Now combine like terms:

==> 3x - x = 5 - 3

==> 2x = 2

**==> x= 1 which is > -1 **

To solve for x:

(3x+3)^1/2 = (x+5)^1/2 .

Square both sides:

3x+3 = x+5

3x-x = 5-3

2x = 2

x = 1

3x+3=x+5 3x-x=5-3 2x=2 x=1