Help - Quadratic in Diophantus I can't understand

[sholto]

I've started reading this fascinating book about the history of complex numbers, but I can't get past page 5! It is describing a passage from Diophantus' Arithmetica and his solution to the problem:

"Given a right-angle triangle of area 7 and perimeter 12, find the the sides."

I can follow as far as: 1/x + 14x +√(1/x² + 196x²) = 12

But then the author says "...which is easily put into the form: 172x = 336x² +24 "

I thought my algebra was pretty good, but I've spent a couple of hours trying to get from the first equation to the second, and I can't do it!

Would someone be kind enough to explain what I'm missing here...

Thankyou

 

[stapel]

I've started reading this fascinating book about the history of complex numbers, but I can't get past page 5! It is describing a passage from Diophantus' Arithmetica and his solution to the problem:

"Given a right-angle triangle of area 7 and perimeter 12, find the the sides."

I can follow as far as: 1/x + 14x +√(1/x² + 196x²) = 12

I suspect the origin of the above equation was something along the lines of the following:

. . . . .\(\displaystyle \mbox{Let the height be indicated by }\, h\, \mbox{ and the base by }\, b. \, \mbox{ Then:}\)

. . . . .\(\displaystyle \dfrac{1}{2}\, bh\, =\, 7\, \mbox{ so }\, bh\, =\, 14\)

. . . . .\(\displaystyle h\, +\, b\, +\, \sqrt{\strut\, h^2\, +\, b^2\,}\, =\, 12\)

. . . . .\(\displaystyle \mbox{Let }\, h\, =\, \dfrac{1}{x}.\, \mbox{Then:}\)

. . . . .\(\displaystyle (b)\,\left(\dfrac{1}{x}\right)\, =\, 14\, \mbox{ so }\, b\, =\, 14x\)

. . . . .\(\displaystyle \dfrac{1}{x}\, +\, 14x\, +\, \sqrt{\strut\, \left(\dfrac{1}{x}\right)^2\, +\, (14x)^2\,}\, =\, 12\)

But then the author says "...which is easily put into the form: 172x = 336x² +24 "

Let's start with the obvious first step: isolating the square root, so that we can then square both sides to eliminate the radical:

. . . . .\(\displaystyle \sqrt{\strut\, \left(\dfrac{1}{x}\right)^2\, +\, (14x)^2\,}\, =\,12\, -\, \dfrac{1}{x}\, -\, 14x\)

. . . . .\(\displaystyle \left(\,\sqrt{\strut\, \left(\dfrac{1}{x}\right)^2\, +\, (14x)^2\,}\,\right)^2\, =\,\left(\, 12\, -\, \dfrac{1}{x}\, -\, 14x\,\right)^2\)

. . . . .\(\displaystyle \dfrac{1}{x^2}\, +\, 196x^2\, =\, -336x\, +\, 172\, +\, 196x^2\, -\, \dfrac{24}{x}\, +\, \dfrac{1}{x^2}\)

. . . . .\(\displaystyle 0\, =\, -336x\, +\, 172\, -\, \dfrac{24}{x}\)

. . . . .\(\displaystyle 0\, =\, -336x^2\, +\, 172x\, -\, 24\)

Move things around a bit, and there you are!

Would someone be kind enough to explain what I'm missing here...

 

[Deleted member 4993]

There is NO real solutions to that equation.

That is why it is included in history of complex numbers - I suppose....

 

[sholto]

Thankyou!

Thankyou Stapel, that's actually fairly straightforward - I was being a dimwit and forgetting expand the terms on the right correctly...

 

[sholto]

Sure Denis.

Book is "An Imaginary Tale: The Story of √-1" by Paul J Nahin

So from right triangle with area 7 and perimeter 12 he starts with:

ab = 14

a + b + √(a² + b²) = 12

Diophantus then reduces the number of variables by observing: a = 1/x and b = 14x (from b = 14/a, I presume)

which gives 1/x + 14x + √(1/x² +196x²) = 12 which is the first equation I quoted.

So many algebra solutions seems to come down to a clever (or inspired) substitution!