Question

solve for x. check for extraneous solutions. $sqrt{3x - 8} + 1 = sqrt{x + 5}$

Explanation:

Step1: Isolate one square root

$\sqrt{3x - 8} = \sqrt{x + 5} - 1$

Step2: Square both sides

$(\sqrt{3x - 8})^2 = (\sqrt{x + 5} - 1)^2$
$3x - 8 = (x + 5) - 2\sqrt{x + 5} + 1$

Step3: Simplify the equation

$3x - 8 = x + 6 - 2\sqrt{x + 5}$
$2x - 14 = -2\sqrt{x + 5}$
$x - 7 = -\sqrt{x + 5}$

Step4: Square both sides again

$(x - 7)^2 = (-\sqrt{x + 5})^2$
$x^2 - 14x + 49 = x + 5$

Step5: Rearrange to quadratic form

$x^2 - 15x + 44 = 0$

Step6: Factor the quadratic

$(x - 4)(x - 11) = 0$
$x = 4 \text{ or } x = 11$

Step7: Check for extraneous solutions

For $x=4$:
$\sqrt{3(4)-8}+1 = \sqrt{4}+1 = 2+1=3$; $\sqrt{4+5}=\sqrt{9}=3$ (valid)
For $x=11$:
$\sqrt{3(11)-8}+1 = \sqrt{25}+1 = 5+1=6$; $\sqrt{11+5}=\sqrt{16}=4$ (6≠4, extraneous)

Answer:

$x=4$