A basic root numbers question

Question

If $\sqrt{x^2+5} - \sqrt{x^2-3} = 2$, then what is $\sqrt{x^2+5} + \sqrt{x^2-3}$?

Answer 1

Notice, we have $$\sqrt{x^2+5} - \sqrt{x^2-3} = 2\tag 1$$ let $$\sqrt{x^2+5} + \sqrt{x^2-3} = y\tag 2$$ Now, multiplying both (1) & (2), we get $$(\sqrt{x^2+5} - \sqrt{x^2-3} )(\sqrt{x^2+5} + \sqrt{x^2-3})=2y$$ $$(\sqrt{x^2+5})^2-(\sqrt{x^2-3})^2=2y$$ $$x^2+5-(x^2-3)=2y$$ $$8=2y\implies y=4$$ Hence, $$\bbox[5px, border:2px solid #C0A000]{\sqrt{x^2+5} + \sqrt{x^2-3} = 4}$$

Answer 2

Hint1:

$$(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$$

Hint2: $$x^{2}+5-(x^{2}-3)=8$$

Answer 3

$$\sqrt{x^2+5} - \sqrt{x^2-3} = 2\\ \sqrt{x^2+5} = \sqrt{x^2-3} + 2\\ (\sqrt{x^2+5})^2 = (\sqrt{x^2-3})^2 + 2^2\\x^2+5=x^2-3+4-4\sqrt{x^2-3} \\5-1=-4\sqrt{x^2-3}\\\sqrt{x^2-3}=1 \\x^2-3=1\\ \rightarrow x=\pm 2\\\sqrt{x^2+5} + \sqrt{x^2-3}=\sqrt{(\pm2)^2+5} - \sqrt{(\pm2)^2-3}=3+1=4$$

Answer 4

We have

$$\sqrt{x^2+5} - \sqrt{x^2-3} = 2 \sqrt{x^2+5} = 2+ \sqrt{x^2-3}\\ x^2+5 = 4+ x^2-3+4\sqrt{x^2-3}\\ 4 = 4\sqrt{x^2-3}\\ 1 = \sqrt{x^2-3}\\ 1 = x^2 -3\\ x^2 = 4\\ x = \pm 2 $$

So that

$$\sqrt{x^2+5} + \sqrt{x^2-3} = 3 + 1 = 4$$