How do you solve #sqrt(6x+5) = sqrt53# and find any extraneous solutions?

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Answer 1 · 2016-09-25T09:56:36

#x=8#

First, we need to square both sides of the equation so that we can easily work with it.

#sqrt(6x+5)^2=sqrt(53)^2#

Doing this will turn the equation into something much more simple.

#6x+5=53#

Next, we subtract #5# from both sides.

#6x=48#

Then, we divide both sides of the equation by #6#.

#x=8#

In order to determine whether or not #x=8# is extraneous, we need to plug it into the original equation.

#sqrt(6x+5)=sqrt(53)#
#sqrt(6(8)+5)=sqrt(53)#
#sqrt(48+5)=sqrt(53)#
#sqrt(53)=sqrt(53)#

Since the statement that resulted in plugging in #x=8# is true, we can confirm that #x=8# is not an extraneous solution.