Question

Question #7a744

Answer 1

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.