Question #7a744

1 Answer
Nov 16, 2016

Answer:

Please see the explanation for steps leading to the answer, #x = 12#

Explanation:

Because the argument of a square root cannot be negative add the restriction #x >= 4# to the original equation:

#sqrt(1 + 2x) - sqrt(2x - 8) = 1; x >=4#

Add #sqrt(2x - 8)# to both sides:

#sqrt(1 + 2x) = sqrt(2x - 8) + 1; x >=4#

Square both sides:

#(sqrt(1 + 2x))^2 = (sqrt(2x - 8) + 1)^2; x >=4#

Squaring "undoes" the radicals on the left and use the pattern #(a + b)^2 = a^2 + 2ab + b^2 on the right:

#1 + 2x = (sqrt(2x - 8))^2 + 2sqrt(2x - 8) + 1; x >=4#

Remove the squared radical on the right:

#1 + 2x = 2x - 8 + 2sqrt(2x - 8) + 1; x >=4#

Combine like terms:

#8 = 2sqrt(2x - 8); x >=4#

Divide both sides by 2:

#4 = sqrt(2x - 8); x >=4#

Square both sides:

#16 = 2x - 8; x >=4#

#2x = 24; x >=4#

#x = 12#

Check #x = 12# in the original equation:

#sqrt(1 + 2(12)) - sqrt(2(12) - 8) = 1#

#sqrt(25) - sqrt(16) = 1#

#5 - 4 = 1#

#1 = 1#

This checks.