Quote: example of solving a quadratic equation

Quotation

First problem, illustrating the first chapter Divide ten into two parts in such a way that one part multiplied by the other and the product, or result, taken four times, will be equal to the product of one part by itself [i.e., 4x(10-x)=x^2; 5x^2=40x; x=8]. The method is to let x represent one part of ten, and the other 10-x. Therefore multiple x by 10-x, giving 10x-x^2. Also multiply 10x-x^2 by 4, as it was to be taken four times, giving 40x-4x^2 as four times the product of one part by the other. Then multiply x by x, i.e. one part by itself, giving x^2, which equals 40x-x^ 2. Therefore restore or complete the number, i.e. add four squares to one square, and you obtain five squares equal to 40x. Hence 8 is the root of the square which itself is 64. The root of this is that part of 10 which is to be multiplied by itself, and the difference between this number and ten is 2. So that 2 is the other part of 10. Now this problem has led you to one of the six chapters and, indeed, to that one in which we treat the type, squares equal to roots.   Google-1   Google-2

Published before 1923

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Topic: history of mathematics (57 items)