2230.51 – Jessamine Bush

Here is a problem proposed by ancient Hindu mathematicians:

"The square root of half the number of bees in a swarm has flown out upon a jessamine bush. Besides these bees, one female bee flies about a male that is buzzing within a lotus flower into which he was allured in the night by its sweet odor and is now imprisoned in it.  Aside from the bees mentioned so far, 89\dfrac{8}{9} of the whole swarm has remained behind. Tell me the number of the bees."

Solution

If xx is the number of bees in the swarm, x2\sqrt{\dfrac{x}{2}} are on the jessamine bush, 2 are at the lotus flower, and 8x9\dfrac{8x}{9} remained behind. So

x2+8x9+2=xx2=x9−2x2=(x9−2)2=x281−4x9+4\begin{align*}\sqrt{\dfrac{x}{2}}+\dfrac{8x}{9}+2&=x \\[1em]\sqrt{\dfrac{x}{2}} &= \dfrac{x}{9}-2 \\[1em]\dfrac{x}{2} &= \left(\dfrac{x}{9}-2\right)^2 \\[1em]&= \dfrac{x^2}{81}-\dfrac{4x}{9}+4\end{align*}

by a spectacular guess, you might multiply through by 162 and get

162x2=162x281−4⋅162x9+4⋅16281x=2x2−72x+6480=2x2−153x+6480=(2x−9)(x−72)\begin{align*}\dfrac{162x}{2}&=\dfrac{162x^2}{81}-\dfrac{4\cdot162x}{9}+4\cdot 162 \\[1em]81x &= 2x^2-72x + 648 \\[0.5em]0 &= 2x^2 -153x +648 \\[0.5em]0 &= (2x-9)(x-72)\end{align*}

So, either by the above or the quadratic formula, x=72x=72 or x=92x = \dfrac{9}{2}, which is not an integer. So x=72x = 72.

Check: x2+8x9+2=6+64+2=72\sqrt{\dfrac{x}{2}}+\dfrac{8x}{9}+2=6 + 64 +2 = 72.