# What is Square root of x^2+4?

Jun 3, 2015

There seem to me to be two aspects to this question:

(1) What does "square root of ${x}^{2} + 4$" mean?

$\sqrt{{x}^{2} + 4}$ is a term which when squared yields ${x}^{2} + 4$ :

$\sqrt{{x}^{2} + 4} \times \sqrt{{x}^{2} + 4} = {x}^{2} + 4$

In other words $t = \sqrt{{x}^{2} + 4}$ is the solution $t$ of the
equation ${t}^{2} = {x}^{2} + 4$

(2) Can the formula $\sqrt{{x}^{2} + 4}$ be simplified?

No.

For starters $\left({x}^{2} + 4\right) > 0$ for all $x \in \mathbb{R}$, so it has no linear factors with real coefficients.

Suppose you produced some formula $f \left(x\right)$ for $\sqrt{{x}^{2} + 4}$. Then $f \left(1\right) = \sqrt{5}$ and $f \left(2\right) = \sqrt{8} = 2 \sqrt{2}$.

So any such formula $f \left(x\right)$ would involve square roots or fractional exponents or suchlike, and be as complex as the original $\sqrt{{x}^{2} + 4}$