Question 54627
{{{sqrt(x+7) + sqrt(x+2) = sqrt(x-1) - sqrt(x-2)}}}
Squaring both sides
{{{(sqrt(x+7) + sqrt(x+2))^2 = (sqrt(x-1) - sqrt(x-2))^2}}}
{{{(x+7)+(x+2) + 2*sqrt(x+7)*sqrt(x+2) = (x-1) + (x-2) - 2*sqrt(x-1)*sqrt(x-2)}}}
{{{2x + 9 + 2*sqrt(x+7)*sqrt(x+2) = 2x - 3 - 2*sqrt(x-1)*sqrt(x-2)}}}
{{{12 + 2*sqrt(x+7)*sqrt(x+2) = - 2*sqrt(x-1)*sqrt(x-2)}}}
{{{6 = -sqrt(x+7)*sqrt(x+2) - sqrt(x-1)*sqrt(x-2)}}}
Squaring again
{{{36 = (x+7)(x+2) + (x-1)*(x-2) + 2*sqrt(x+7)*sqrt(x+2)*sqrt(x-1)*sqrt(x-2)}}}
{{{36 = x^2+9x+14 + x^2-3x+2 + 2*sqrt(x+7)*sqrt(x+2)*sqrt(x-1)*sqrt(x-2)}}}
{{{36 = 2x^2+6x+16 + 2*sqrt(x+7)*sqrt(x+2)*sqrt(x-1)*sqrt(x-2)}}}
{{{10 = x^2+3x + sqrt(x+7)*sqrt(x+2)*sqrt(x-1)*sqrt(x-2)}}}
{{{x^2 - 3x - 10= -sqrt(x+7)*sqrt(x+2)*sqrt(x-1)*sqrt(x-2)}}}
{{{(x+2)(x-5) = -sqrt(x+7)*sqrt(x+2)*sqrt(x-1)*sqrt(x-2)}}}
Squaring again
{{{(x+2)^2(x-5)^2 = (x+7)(x+2)(x-1)(x-2)}}}
{{{(x+2)((x+2)(x-5)^2 - (x+7)(x-1)(x-2)) = 0}}}
{{{(x+2)((x+2)(x^2 + 10x + 25) - (x^2+6x-7)(x-2)) = 0}}}
{{{(x+2)(x^3 + 10x^2 + 25x + 2x^2 + 20x + 50 - x^3-6x^2+7x +2x^2+12x-14) = 0}}}
{{{(x+2)(8x^2 + 64x + 36) = 0}}}
{{{(x+2)(2x^2 + 16x + 9) = 0}}}

Clearly no root of this equation is equal to 2. The roots are {{{x = (-16 +- sqrt( 16^2-4*2*9 ))/(2*2)}}} & -2 i.e. {{{x = -4 +- sqrt(46)/2}}} & -2.

So all values of the x are negative. So the given equation is no longer valid in real domain.

Note that the equation is satisfied by x=2 if it is {{{sqrt(x+7) - sqrt(x+2)=...}}} and not {{{sqrt(x+7) + sqrt(x+2)=...}}}