How do you solve #sqrt(6-b)=sqrt(b+10)# and check your solution?

1 Answer
Jul 25, 2017

Answer:

See a solution process below:

Explanation:

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

#(sqrt(6 - b))^2 = (sqrt(b + 10))^2#

#6 - b = b + 10#

Next, add #color(red)(b)# and subtract #color(blue)(10)# from each side of the equation to isolate the #b# term while keeping the equation balanced:

#6 - b + color(red)(b) - color(blue)(10) = b + 10 + color(red)(b) - color(blue)(10)#

#6 - color(blue)(10) - b + color(red)(b) = b + color(red)(b) + 10 - color(blue)(10)#

#-4 - 0 = 1b + color(red)(1b) + 0#

#-4 = (1 + color(red)(1))b#

#-4 = 2b#

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

#-4/color(red)(2) = (2b)/color(red)(2)#

#-2 = (color(red)(cancel(color(black)(2)))b)/cancel(color(red)(2))#

#-2 = b#

#b = -2#

To validate the answer we will substitute #color(red)(-2)# for each occurrence of #color(red)(b)# in the original equation and determine if both sides of the equation are equal:

#sqrt(6 - color(red)(b)) = sqrt(color(red)(b) + 10)# becomes:

#sqrt(6 - color(red)(-2)) = sqrt(color(red)(-2) + 10)#

#sqrt(6 + color(red)(2)) = sqrt(color(red)(-2) + 10)#

#sqrt(8) = sqrt(8)#

Both sides of the equation are equal so the answer is correct.