Question 606659
Solve square root of 2x+3- square root of x+1=1
:
{{{sqrt(2x+3)-sqrt(x+1) = 1}}}
:
{{{sqrt(2x+3) = sqrt(x+1) + 1}}}
Square both sides, you have to FOIL the right side
2x + 3 = {{{(x+1) + 2sqrt(x+1) + 1}}}
:
2x + 3 = x + 2 + {{{2sqrt(x+1)}}}
:
2x - x + 3 - 2 = {{{2sqrt(x+1)}}}
:
(x + 1) = {{{2sqrt(x+1)}}}
:
Square both sides, FOIL the left side
:
x^2 + 2x + 1 = 4(x+1)
x^2 + 2x + 1 = 4x + 4
x^2 + 2x - 4x + 1 - 4 = 0
x^2 - 2x - 3 = 0
Factors to
(x-3)(x+1) = 0
Two solutions
x = 3
x = -1
;
We have to check both solutions in the original equation
x = -1 will not work, it would give us a neg value inside the radical
Try x = 3
{{{sqrt(6+3)-sqrt(3+1) = 1}}}; looks OK, x=3 is our solution