# Question 7a744

Nov 16, 2016

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 $x \ge 4$ to the original equation:

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

Add $\sqrt{2 x - 8}$ 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:

$\sqrt{1 + 2 \left(12\right)} - \sqrt{2 \left(12\right) - 8} = 1$

$\sqrt{25} - \sqrt{16} = 1$

$5 - 4 = 1$

$1 = 1$

This checks.