Question

solving equations containing two radicals
solving equations containing two radicals
after eliminating radicals, what quadratic equation can you solve to find the potential solutions of $sqrt{2x+3}-sqrt{x+1}=1$
$x^2 - 2x - 3 = 0$
$x^2 + 2x + 5 = 0$
$x^2 - 4x - 3 = 0$
$2x^2 - 4x - 4 = 0$

Explanation:

Step1: Isolate one radical

$\sqrt{2x+3} = \sqrt{x+1} + 1$

Step2: Square both sides

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

Step3: Simplify and isolate remaining radical

$2x+3 = x+2 + 2\sqrt{x+1}$
$x+1 = 2\sqrt{x+1}$

Step4: Square both sides again

$(x+1)^2 = (2\sqrt{x+1})^2$
$x^2 + 2x + 1 = 4(x+1)$

Step5: Simplify to quadratic

$x^2 + 2x + 1 = 4x + 4$
$x^2 - 2x - 3 = 0$

Answer:

$x^2 - 2x - 3 = 0$