# Determine the root of the equation (3x + 5)^1/2 = 5 .Determine the root of the equation (3x + 5)^1/2 = 5 .

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### 6 Answers

You need to raise to square both sides such that:

You need to verify this value of x in equation such that:

Hence, evaluating the solution to equation yields .

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

First let us square both sides to get rid of the root:

==> (3x + 5 + 5^2

==> 3x + 5 = 25

Now subtract 5 from both sides:

==> 3x + 20

Now divide by 3:

**==> x= 20/3**

The root of an equation is where the graph of the equation crosses the x-axis. The equation that you have is not in the most obvious form to determine its root as once simplified we will see the equation is in the form of x = n where n is the root:

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

3x+5 = 5^2 = 25

3x = 25-5 = 20

x = 20/3

Therefore the root of this equation is x = 20/3 assuming that 3x+5 > 0 as you cannot square root a negative number. Otherwise this equation would be undefined...

We have to find the root of (3x+5)^(1/2)=5.

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

=> (3x+5) = 5^2

=> (3x+5) = 25

=> 3x = 25-5

=> 3x = 20

=> x = 20/3

For x =20/3, 3x+5= 25 which is positive.

**Therefore x= 20/3**

To find the root of (3x+5)^(1/2) = 5.

Square both sides.

3x+5 = 5^2 = 25

Subtract 5.

3x = 25-5 = 20

3x= 20. Divide by 3.

x = 20/3

Before solving the equation, we'll impose constraint of existence of the square root, namely 3x + 5>0.

We'll solve the inequality:

3x + 5>0

We'll subtract 5 both sides:

3x>-5

We'll divide by 3:

x>-5/3

The interval of admissible values for x is (-5/3 , +inf.)

Now, we'll eliminate the square root. For this reason, we'll raise to square, both sides of the equation:

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

We'll eliminate the square root from the left side :

3x + 5 = 25

We'll subtract 5 both sides:

3x = 25-5

3x = 20

We'll divide by 3 both sides:

**x = 20/3**

**Because 20/3 belongs to the interval of admissible values, the solution is valid. So, the equation has the solution x = 20/3.**