How do you solve #sqrt(8x)+1=65# and check the solution?

1 Answer
May 23, 2017

Answer:

See a solution process below:

Explanation:

First, subtract #color(red)(1)# from each side of the equation to isolate the radical while keeping the equation balanced:

#sqrt(8x) + 1 - color(red)(1) = 65 - color(red)(1)#

#sqrt(8x) + 0 = 64#

#sqrt(8x) = 64#

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

#(sqrt(8x))^2 = 64^2#

#8x = 4096#

Now, divide each side of the equation by #color(red)(8)# to solve for #x# while keeping the equation balanced:

#(8x)/color(red)(8) = 4096/color(red)(8)#

#(color(red)(cancel(color(black)(8)))x)/cancel(color(red)(8)) = 512#

#x = 512#

To check the solution substitute #color(red)(512)# for #color(red)(x)# and calculate each side of the equation to ensure both sides are equal:

#sqrt(8color(red)(x)) + 1 = 65# becomes:

#sqrt(8 * color(red)(512)) + 1 = 65#

#sqrt(4096) + 1 = 65#

#64 + 1 = 65#

#65 + 65#

Both sides of the equation are equal therefore the solution is valid.