How do you solve #sqrt(6x-20)=4# and find any extraneous solutions?

1 Answer
Sep 10, 2017

Answer:

See a solution process below:

Explanation:

First, square both sides of the equation to eliminate the square root while keeping the equation balanced:

#(sqrt(6x - 20))^2 = 4^2#

#6x - 20 = 16#

Next, add #color(red)(20)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#6x - 20 + color(red)(20) = 16 + color(red)(20)#

#6x - 0 = 36#

#6x = 36#

Now, divide each side of the equation by #color(red)(6)# to solve for #x# while keeping the equation balanced:

#(6x)/color(red)(6) = 36/color(red)(6)#

#(color(red)(cancel(color(black)(6)))x)/cancel(color(red)(6)) = 6#

#x = 6#

If we substitute #color(red)(6)# for #color(red)(x)# in the original equation and evaluate the square root we will find the extraneous solutions. Remember, the square root of a number produces a positive and negative result:

#sqrt(6color(red)(x) - 20) = 4# becomes:

#+-sqrt((6 * color(red)(6)) - 20) = 4#

Or

#-sqrt((6 * color(red)(6)) - 20) = 4# and #sqrt((6 * color(red)(6)) - 20) = 4#

#-sqrt(36 - 20) = 4# and #sqrt(36 - 20) = 4#

#-sqrt(16) = 4# and #sqrt(16) = 4#

#-4 != 4# and #4 = 4#

#-sqrt(16)# is an extraneous solution.