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ted geometrically; but the geometrical construction of problems may be effected also, without resolving the equation, and even without deducing a final equation, by the methods afterwards to be explained.
If the final equation is simple or quadratic, the roots being obtained by the common rules, may be geometrically exhibited by the finding of proportionals, and the addition or subtraction of squares.
By inserting numbers for the known quantities, a numeral expression of the quantities sought will be obtained by resolving the equation. But, in order to determine some particulars of the problem, bet fides finding the unknown quantities of the equation, it may be farther necessary to make a simple construction ; or, if it is required that every thing be expressed in numbers, to substitute a new calculation in place of that construction.
To divide a given straight line AB into two
parts, so that the rectangle contained by the whole line, and one of the parts, may be equal to the square of the other part.
This is Prop. IIth II. B. of Euclid.
Let C be the point of division, and let AB=a, AC=x, and then CB=a-x. From the problem a?-ax=x?; and this equation being resolved (Chap. V. P. II.) gives x=+
V The quantity Vata, is the hypothenuse of a right angled triangle of which the two sides are a, and, and is therefore
a easily found; being taken from this line, gives x=AC, which is the
folution. But if a line Ac be taken on the opposite
side of A, and equal to the above mentioned hypothenuse, together with it will
represent the negative root - a
This solution coincides with what is gi-
=EF-EA=AF=AH; and the point H corresponds to C in the preceding figure.
Besides, if on EF+EA=CF (instead of EF-EA=FA) a fquare be described on the opposite side of CF from AG, BA produced will meet a fide of it in a point, which if it be called K, will give KBX BA
=KA’ K corresponds to c; and this folution will correspond with the algebraical solution, by means of the negative root.
a? a? +
· If CB had been called x, and AC=ax,
=X the equation would be ax=a?-2ax+xa, which gives xa
3a&v5a, in which both roots are positive, and the solutions derived from them coincide with the preceding. If the solution be confined to a point within the line, then one of the positive roots must be rejected, for one of the roots of the compound square from which it is derived is *--- 30, a negative quantity, which in this strict hypothesis is not admitted. In such a problem, however, both constructions are generally received, and considered even as necessary to a complete solution of it.
If a solution in numbers be required, let AB=10, then x==125-5. It is plain, whatever be the value of AB, the roots of this equation are incommensurate, though they may be found, by approximation, to any degree of exactness required. In this case, x= £11.1803-5, nearly, that is AC=6.1803 nearly; and Ac=16.1803 nearly.
In a given Triangle ABC to infcribe a Square.
Suppose it to be done, and let it be EFHG; from A let AD be perpendicular on the basé BC, meeting EF in K.
B square can easily be
G D H constructed, and let it be called x: Then (KD=EG=)EF=P
On account of the parallels EF, BC, AD:BC:: AK : EF; that is, p:a:: x:P-X, and pa-px=ax, which equation being resolved, gives x=
Therefore x or AK is a third proportional to pta, and p, and may be found by 11. VI. El. The point K being found, the construction of the square is sufficiently obvious.