Question

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

Answer 1

$\sqrt{x+1}+ \sqrt{3-x} =2\\ \\ \sqrt{x+1}=2-\sqrt{3-x} \\ \\ ( \sqrt{x+1})^2=(2-\sqrt{3-x})^2 \\ \\ x+1=4-4\sqrt{3-x}+3-x\\ \\ 2x-6=-4\sqrt{3-x}\\ \\ (2x-6)^2=(-4\sqrt{3-x})^2\\ \\ 4x^2-24x+36=16(3-x)\\ \\ 4x^2-24x+36=48-16x\\ \\ 4x^2-8x-12=0\\ \\ x^2-2x-3=0\\ \\ x_1=-1\\ \\ x_2=3$