How do you solve #sqrt(2x+5) +6=4#?

1 Answer
Aug 11, 2017

Answer:

See a solution process below:

Explanation:

First, subtract #color(red)(6)# from each side of the equation to isolate the radical while keeping the equation balanced:

#sqrt(2x + 5) + 6 - color(red)(6) = 4 - color(red)(6)#

#sqrt(2x + 5) + 0 = -2#

#sqrt(2x + 5) = -2#

Next, square each side of the equation to eliminate the radical while keeping the equation balanced:

#(sqrt(2x + 5))^2 = -2^2#

2x + 5 = 4#

Now, subtract #color(red)(5)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#2x + 5 - color(red)(5) = 4 - color(red)(5)#

#2x + 0 = -1#

#2x = -1#

Now, divide each side of the equation by #color(red)(2)# to solve for #x# while keeping the equation balanced:

#(2x)/color(red)(2) = -1/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = -1/2#

#x = -1/2#

To check the answer we can substitute #-1/2# for #x# and calculate the result:

#sqrt(2x + 5) + 6 = 4# becomes:

#sqrt((2 xx -1/2) + 5) + 6 = 4#

#sqrt(-1 + 5) + 6 = 4#

#sqrt(4) + 6 = 4#

Remember, the square root of a number produces both a positive AND negative result:

#-2 + 6 = 4# and #2 + 6 = 4#

#4 = 4# and #8 != 4#

The solution for #sqrt(4) = 2# is an extraneous solution.

The solution for #sqrt(4) = -2# is a valid solution.