What is Square root of x^2+4?

Question

29408 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-03T19:06:21

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

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

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

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

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

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

No.

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

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

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