How do you solve #sqrt(2n-88)=sqrt(n/6)#?

1 Answer
Oct 31, 2016

Answer:

The solution of the equation is #n = 48#.

Explanation:

First, you must get rid of the radicals. In this case, that is achieved by squaring both sides of the equation.

#(sqrt(2n - 88))^2 = (sqrt(n/6))^2#

#2n - 88 = n/6#

Now use inverse operations to solve for #n#.

#(2n - 88)6 = (n/6)6#

#12n - 528 = n#
#12n - 12n - 528 = n - 12n#
#-528 = -11n#
#(-528)/-11 = (-11n)/-11#

#48 = n#

Now we must check the solution to be sure that it is not extraneous. This step absolutely cannot be skipped when solving radical equations! Remember to check a solution, we put it in place of the variable in the original equation and simplify each side of the equation to see if it makes a true statement.

#sqrt(2*48 - 88) = sqrt(48/6)#

#sqrt(96 - 88) = sqrt(8)#
#sqrt(8) = 2sqrt(2)#
#2sqrt(2) = 2sqrt(2)#

Since the two sides of the equation simplify to be equal, #48# is the solution of the equation.