# Solve the radical equation. \sqrt{x+16}-\sqrt{x-4}=2

Solve the radical equation.
$$\displaystyle\sqrt{{{x}+{16}}}-\sqrt{{{x}-{4}}}={2}$$

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Step 1
$$\displaystyle\sqrt{{{x}+{16}}}-\sqrt{{{x}-{4}}}={2}$$
Step 2
$$\displaystyle\sqrt{{{x}+{16}}}-\sqrt{{{x}-{4}}}={2}$$
$$\displaystyle\Rightarrow\sqrt{{{x}+{16}}}=\sqrt{{{x}-{4}}}+{2}$$
squaring both sides
$$\displaystyle{\left(\sqrt{{{x}+{16}}}\right)}^{{{2}}}={\left(\sqrt{{{x}-{4}}}+{2}\right)}^{{{2}}}$$
Expanding both sides
$$\displaystyle{x}+{16}={x}-{4}+{2}{\left(\sqrt{{{x}-{4}}}\right)}{\left({2}\right)}+{4}\ \ \ {\left[{\left({a}+{b}\right)}^{{{2}}}={a}^{{{2}}}+{2}{a}{b}+{b}^{{{2}}}\right]}$$
$$\displaystyle\Rightarrow{x}+{16}={x}+{4}\sqrt{{{x}-{4}}}$$
$$\displaystyle\Rightarrow{16}={4}\sqrt{{{x}-{4}}}$$
$$\displaystyle\Rightarrow{\frac{{{16}}}{{{4}}}}=\sqrt{{{x}-{4}}}$$
$$\displaystyle\Rightarrow{4}=\sqrt{{{x}-{4}}}$$
squaring both sides
16=x-4
$$\displaystyle\Rightarrow{20}={x}$$