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The solution of the equation √{(x−a)(x−b)}(x−c)(x−d)}=e

Published online by Cambridge University Press:  20 January 2009

J. D. Hamilton Dickson
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If P=a+b−c−d, and K=cd−ab, the solution of the equation

may be most readily found by putting (x−a)(x−b)=z2; whence (x−c)(x−d)=z2+Px+K, and therefore =z2−2ez+e2, and finally x is given by the equation

Type
Research Article
Information
Proceedings of the Edinburgh Mathematical Society , Volume 9 , February 1890 , pp. 13 - 23
Copyright
Copyright © Edinburgh Mathematical Society 1890