Here is a problem allegedly proposed by ancient Hindu mathematicians: The square root of half the number of bees in a swarm has flown out upon a jessamine bush; 8/9 of the whole swarm has remained behind; and one female bee flies about a male that is buzzing within a lotus flower into which he was lured in the night by its sweet scent, but is now imprisoned in it. Tell me the number of the bees in the swarm.
Let x be the number of bees in the swarm.
Then √x/2 are on the jessamine bush,
8x/9 remain behind,
and 2 are at the lotus flower.
8x/9 remain behind,
and 2 are at the lotus flower.
So,
(√x/2) + (8x/9) + 2 = x
√x/2 = x/9 – 2
x/2 = (x/9 – 2)2
x/2 = x2/81 – 4x/9 + 4
(x/2) · 162 = (x2/81 – 4x/9 + 4) · 162
81x = 2x2 – 72x + 648
2x2 – 153x + 648 = 0
(2x – 9)(x – 72) = 0
x = 72 or 9/2
√x/2 = x/9 – 2
x/2 = (x/9 – 2)2
x/2 = x2/81 – 4x/9 + 4
(x/2) · 162 = (x2/81 – 4x/9 + 4) · 162
81x = 2x2 – 72x + 648
2x2 – 153x + 648 = 0
(2x – 9)(x – 72) = 0
x = 72 or 9/2
So x = 72.