How do you solve sqrt(6x+5) = sqrt536x+5=53 and find any extraneous solutions?

1 Answer
Sep 25, 2016

x=8x=8

Explanation:

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

sqrt(6x+5)^2=sqrt(53)^26x+52=532

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

6x+5=536x+5=53

Next, we subtract 55 from both sides.

6x=486x=48

Then, we divide both sides of the equation by 66.

x=8x=8

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

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

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