Why this value is not the second root of the equation?

Question

$\sqrt{x+3}=x-2$

Why $\frac{5}{2} - \frac{\sqrt{21}}{2}$ is not root?

There is only one restriction: $\sqrt{x+3}$, but $\frac{5}{2} - \frac{\sqrt{21}}{2} > 0$.

$x^2-4x+4=x+3$, $x^2-5x+1=0$, $D=25-4=21>0$

$D>0$, =>, $x = \frac{ 5 \pm \sqrt{21} }{ 2 }$

Answer 1

Because $\left(\frac{5}{2} - \frac{\sqrt{21}}{2}\right)-2$ is negative, so it cannot be the square root of something.

Answer 2

There is actually a second restriction to your equation, which is "hidden":

Since $\sqrt{x+3} \geq 0$ you must also have $x-2 \geq 0$, which your root fails.

Note that the second root actually satisfies $\sqrt{x+3}=-(x-2)$.

Answer 3

That is because the irrational equation: $$\sqrt A=B\iff A=B^2\;\textbf{ and }\, B\ge 0,$$ i.e., in the present case, $\;x\ge 2$, whereas $\;\dfrac{5-\sqrt{21}}2<\dfrac12$.