How do you solve and check for extraneous solutions in #3sqrt(2x+4)=12#?

Question

Eli Assouline · Accepted Answer

#x = 6#

First, isolate the radical on one side of the equation by dividing both sides by ##

#(color(red)(cancel(color(black)(3))) * sqrt(2x+4))/color(red)(cancel(color(black)(3))) = 12/3#

#sqrt(2x+4) = 4#

Square both sides of the equation to get rid of the square root

#(sqrt(2x+4))^2= 4^2#

#2x+4 = 16#

Finally, isolate #x# on one side of the equation by adding #-4# to both sides and dividing everything by #2#

#2x + color(red)(cancel(color(black)(4))) - color(red)(cancel(color(black)(4))) = 16 - 4#

#(color(red)(cancel(color(black)(2)))x)/color(red)(cancel(color(black)(2))) = 12/2#

#x = color(green)(6)#

Eva Abramson · Answer

undefined To solve and check for extraneous solutions in the equation 3√(2x+4) = 12, follow these steps:

1. Start by isolating the radical term by dividing both sides of the equation by 3: √(2x+4) = 4.

2. Square both sides of the equation to eliminate the square root: (2x+4) = 16.

3. Simplify the equation by subtracting 4 from both sides: 2x = 12.

4. Divide both sides of the equation by 2 to solve for x: x = 6.

5. Now, substitute the found value of x back into the original equation to check for extraneous solutions.

For x = 6, the original equation becomes: 3√(2(6)+4) = 12.

Simplifying further: 3√(12+4) = 12, 3√16 = 12, 3(4) = 12, 12 = 12.

Since the equation is true, x = 6 is a valid solution and there are no extraneous solutions.

Therefore, the solution to the equation 3√(2x+4) = 12 is x = 6, and there are no extraneous solutions.