1. ## Ancient mathematics

Hello everyone,

Solve by the method of your choice: 1-third of a school of fish and 3 times the square root of the remaining part of the school were seen in the ocean; and in a river was seen a male fish along with 3 female fish making up the remainder of the school. How many fish were in the school?

My attempt:

I think one would have to use the quadratic equation and one would set it up like the following:

1/3x+3(square root(2/3x))+1+3=x

From here would I solve for x by using the quadratic equation?

Could someone please tell me if this is correct so far?

Thank you very much

2. ## Yes, you're on the right track

x/3 + 3sqrt(2x/3) + 3 + 1 = x

Isolate the square root on the left and clear the denominator on the right hand side by multiplying the 3 that turns up:

3sqrt(2x/3) = (2x-12)/3

-> 9sqrt(2x/3) = 2x-12

now we can square both sides:

81*2x/3 = 4*(x-6)² = 4*(x²-12x+36) = 4x²-48x+144

Collect and put into the form ax^2 + bx + c = 0 which can then be solved in any way you like

4x² - 102x + 144 = 0

3. Thank you very much

I got 24 and 1.5 by using the quadratic formula, but since you can't have half a fish, 24 is the only answer. Does that look right?

Also, if you are given an ancient problem do you know the rule for when to use the quadratic formula and when to do it by "inversion" (working backwards)?

For example,

The number which multiplied by 2, added to three-fourths of the result, divided by 6, diminished by 1-third of the result, multiplied by itself, decreased by fifty-two, having its square-root taken, increased by 7, and divided by 12, produces x. Solve for x.

or

Out of a school of fish 1/5 went to place a., one-third went to place b., and three times the difference of those went to place c. One remaining fish, swam in a circle. How many fish were there?

1/5x+1/3x+3(1/3x-1/5x)+1=x

1/5x+1/3x+2/5x+1=x

1-(1/5-1/3-2/5)=1/15

1/1/15=15

Thank you