Question #7a744

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Answer 1 · 2016-11-16T20:19:50

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

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.