Question 188921
Find all real number solutions of the equation algebraically:
sqrt (2x+1) - sqrt (x+1) = 2

Can someone show me how to solve the problem by first isolating sqrt (2x+1) on the left side of the equation and then squaring each side?

So the first step gives us...
sqrt (2x+1) = 2+ sqrt (x+1)
:
Squaring both sides gives you 2x + 1 on the left but you have to FOIL the right side:
2x + 1 = 4 + 2sqrt(x+1) + 2sqrt(x+1) + (sqrt(x+1))^2
which is:
2x + 1 = 4 + 4sqrt(x+1) + x + 1
:
2x + 1 = 5 + x + 4sqrt(x+1)

Isolate the radical by itself on the right
2x - x + 1 - 5 = 4sqrt(x+1)
:
(x - 4) = 4sqrt(x+1)
:
Square both sides again, FOIL the left side
x^2 - 8x + 16 = 16(x+1)
:
x^2 - 8x + 16 = 16x + 16
:
x^2 - 8x - 16x + 16 - 16 = 0
:
x^2 - 24x = 0
Factor out x
x(x - 24) = 0
Two solutions
x = 0
and
x = +24, this is the only solution that works in the original equation
;
:
Did this help?