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(3x+3)^1/2 = (x+5)^1/2    find x

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talaltalool | Student, Grade 10 | eNoter

Posted October 4, 2010 at 1:51 AM via web

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(3x+3)^1/2 = (x+5)^1/2    find x

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giorgiana1976 | College Teacher | Valedictorian

Posted October 4, 2010 at 1:55 AM (Answer #1)

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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.

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted October 4, 2010 at 1:52 AM (Answer #2)

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(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

 

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neela | High School Teacher | Valedictorian

Posted October 4, 2010 at 1:56 AM (Answer #3)

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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

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phucnguyen08 | Student, College Freshman | Honors

Posted October 4, 2010 at 2:40 AM (Answer #4)

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

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