Calculation:

Consider the equation, \(\displaystyle\sqrt{{{x}+{4}}}-{2}={x}\)

\(\displaystyle\sqrt{{{x}+{4}}}-{2}={x}\)

\(\displaystyle\sqrt{{{x}+{4}}}={x}+{2}\)

Now, squaring both sides,

\(\displaystyle{\left(\sqrt{{{x}+{4}}}\right)}^{{{2}}}={\left({x}+{2}\right)}^{{{2}}}\)

\(\displaystyle{x}+{4}={\left({x}\right)}^{{{2}}}+{\left({2}\right)}^{{{2}}}+{2}{\left({x}\right)}{\left({2}\right)}\)

\(\displaystyle{x}+{4}={x}^{{{2}}}+{4}{x}+{4}\)

Further simplify and taking x common from left side of the equation,

\(\displaystyle{x}^{{{2}}}+{3}{x}={0}\)

\(\displaystyle{x}{\left({x}+{3}\right)}={0}\)

Now, apply zero product rule,

\(\displaystyle{x}{\left({x}+{3}\right)}={0}\)

\(\displaystyle{x}={0}\) or \(\displaystyle{\left({x}+{3}\right)}={0}\)

\(\displaystyle{x}={0}\) or \(\displaystyle{x}=-{3}\)

Hence, the solution of the equation \(\displaystyle\sqrt{{{x}+{4}}}-{2}={x}\) and \(\displaystyle{x}={0}\) or \(\displaystyle{x}=-{3}\).

Now, to check the solution put the values of x in the original equation,

Substitute \(\displaystyle{x}={0}\) in the equation \(\displaystyle\sqrt{{{x}+{4}}}-{2}={x}\)

\(\displaystyle\sqrt{{{x}+{4}}}-{2}={x}\)

\[\sqrt{\left(0\right)+4}-2\left(0\right)\]

\[2-20\]

\(\displaystyle{0}={0}\)

Since, left side is equal to right side so the solution is true.

Now, substitute \(\displaystyle{x}=-{3}\) in the equation \(\displaystyle\sqrt{{{x}+{4}}}-{2}={x}\)

\(\displaystyle\sqrt{{{x}+{4}}}-{2}={x}\)

\[\sqrt{\left(-3\right)+4}-2\left(-3\right)\]

\[1-2-3\]

\(\displaystyle-{1}=-{3}\)

Here, left side is not equal to right side so the solution is false.

Therefore, the solution set is (0) and the value \(\displaystyle{x}=-{3}\) does not check.

Consider the equation, \(\displaystyle\sqrt{{{x}+{4}}}-{2}={x}\)

\(\displaystyle\sqrt{{{x}+{4}}}-{2}={x}\)

\(\displaystyle\sqrt{{{x}+{4}}}={x}+{2}\)

Now, squaring both sides,

\(\displaystyle{\left(\sqrt{{{x}+{4}}}\right)}^{{{2}}}={\left({x}+{2}\right)}^{{{2}}}\)

\(\displaystyle{x}+{4}={\left({x}\right)}^{{{2}}}+{\left({2}\right)}^{{{2}}}+{2}{\left({x}\right)}{\left({2}\right)}\)

\(\displaystyle{x}+{4}={x}^{{{2}}}+{4}{x}+{4}\)

Further simplify and taking x common from left side of the equation,

\(\displaystyle{x}^{{{2}}}+{3}{x}={0}\)

\(\displaystyle{x}{\left({x}+{3}\right)}={0}\)

Now, apply zero product rule,

\(\displaystyle{x}{\left({x}+{3}\right)}={0}\)

\(\displaystyle{x}={0}\) or \(\displaystyle{\left({x}+{3}\right)}={0}\)

\(\displaystyle{x}={0}\) or \(\displaystyle{x}=-{3}\)

Hence, the solution of the equation \(\displaystyle\sqrt{{{x}+{4}}}-{2}={x}\) and \(\displaystyle{x}={0}\) or \(\displaystyle{x}=-{3}\).

Now, to check the solution put the values of x in the original equation,

Substitute \(\displaystyle{x}={0}\) in the equation \(\displaystyle\sqrt{{{x}+{4}}}-{2}={x}\)

\(\displaystyle\sqrt{{{x}+{4}}}-{2}={x}\)

\[\sqrt{\left(0\right)+4}-2\left(0\right)\]

\[2-20\]

\(\displaystyle{0}={0}\)

Since, left side is equal to right side so the solution is true.

Now, substitute \(\displaystyle{x}=-{3}\) in the equation \(\displaystyle\sqrt{{{x}+{4}}}-{2}={x}\)

\(\displaystyle\sqrt{{{x}+{4}}}-{2}={x}\)

\[\sqrt{\left(-3\right)+4}-2\left(-3\right)\]

\[1-2-3\]

\(\displaystyle-{1}=-{3}\)

Here, left side is not equal to right side so the solution is false.

Therefore, the solution set is (0) and the value \(\displaystyle{x}=-{3}\) does not check.