Question

Find the non-extraneous solutions of the square root x+6-5=x+1

Answer 1

$\sqrt{x+6}-5=x+1\ \ \ |+5\\\sqrt{x+6}=x+6\\\\\text{The domain:}\\x+6\geq0\to x\geq-6\\\\D:x\geq-6\\\\\sqrt{x+6}=x+6\ \ \ |\text{square both sides}\\x+6=(x+6)^2\ \ \ |\text{Use}\ (a+b)^2=a^2+2ab+b^2\\x+6=x^2+12x+36\ \ \ |-x-6\\x^2+6x+30=0\\\\\text{Use the quadratic formula}\\\\a=1;\ b=6;\ c=30\\\\b^2-4ac=6^2-4\cdot1\cdot30=36-120=-84 \ \textless \ 0\\\\\text{No real solutions}$