Question

\[\sqrt{2x+3}-\sqrt{x+1}=1\]

Answer 1

put radical on either side of = and then squre both sides\[\left( \sqrt{2x+3} \right)^2=\left( 1+\sqrt{x+1} \right)^2\]\[2x+3=1+2\sqrt{x+1}+x+1\]isolate the radical and square again\[(x+1)^2=\left( 2\sqrt{x+1} \right)^2\]\[x^2+2x+1=4x+4\]collect like terms and solve\[x^2-2x-3=0\] \[(x-3)(x+1)=0\]\[x=3 \ or \ x=-1\]these have to be considered possible solutions; each should checked in the equation at a point before the first squaring.

Answer 2

x=3 and the x=-1 both check. Do you need to see?

Answer 3

aha no i got it but thanks soo much! but ima take a break!

Answer 4

later