Can You Solve a Radical Equation?

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[mokouf]

This equation has absolutely no solution, not even in complex plane.

[akashdhandhi]

√(X+2) -√(x+3) =1
√(x+2) =1+√(x+3)
x+2=1+(x+3) +2√(x+3)
0=2+2√(x+3)
-1=√(x+3) so here equal -2 gives us one solution which fails the problem equation and hence no possible solutions.

[zinzhao]

Super beautiful representation of a proof.

[ericerpelding]

If x is allowed to be complex, then one gets 2=3, so no complex solutions either!

[Hobbitangle]

√(x+2)-√(x+3)=1
Let u=√(x+2) ≥ 0, v=√(x+3) ≥ 0
u-v=1
u²=x+2
v²=x+3
Subracting
u²-v²=-1
(u-v)(u+v)=-1
u+v=-1, because of u-v=1

u=0; v=-1
But the value of v doesn't satisfy the constraint v≥0, so there're no roots.

[morteza]

Sqrt(x+2)-sqrt(x+3)=1
Multiply it by its conjugate
-1/(sqrt(x+2)+sqr(x+3))<0 but 0<1

[royalredbird]

The funny thing is that this equation has no complex solution either. Forget about the real ones.

[timewalkwalker]

Well I only know introductory limits (I'm not high schooler ) can't we just use limits since our number system isn't developed enough to solve this equation by using complex plane or real no.