Triangular Roots

While everyone has heard of a "square root", I for one had not heard of a "triangular root" defined for a number \(x\) as:$$ \sqrt [\strut \triangle] {x} = \frac{-1 \pm \sqrt{1+8x}}{2}$$Suppose we want to find the triangular root of 6. This gives:$$  \begin{align} \sqrt [\strut \triangle] {6} &= \frac{-1 \pm \sqrt{1+8 \times 6}}{2}\\  \\ &=  \frac{-1 \pm \sqrt{1+48}}{2} \\  \\ &= \frac{-1 \pm \sqrt{49}}{2} \\  \\ &= \frac{-1 \pm 7}{2}\\ \\ &= 3 \text{ } \text{  or} -4 \end{align}$$The formula arises from the definition of a triangular number \(x\) and the solution of the resulting quadratic equation:$$ \begin{align}  \frac{n \, (n+1)}{2} &= x \\ \\ n^2 + n -2x &=0 \\ \\  n &= \frac{-1 \pm \sqrt{1+8x}}{2} \end{align} $$This method is no different to what we do when finding the square root of a number where we have:$$ \begin{align} n^2 &= x \\ n &= \pm \sqrt{x}  \end{align}$$By way of comparison it can be noted that triangular roots are only real if \(x \geq -1/8 \) whereas square roots are only real if \(x \geq 0 \). I can thank Dr. Barker's YouTube video for prompting this post. 

Figure 1 shows the situation for \(6 = \dfrac{n \, (n+1)}{2}\) where \(n=-4 \text{ or } 3\).

Figure 1:  -4 and 3 are the values of \( \sqrt [\strut \triangle] {6}\)