Part I. Section IV. Chapter 5. “Of the Resolution of Pure Quadratic Equations.”

623 An equation is said to be of the second degree, when it contains the square, or the second power, of the unknown quantity, without any of its higher powers; and an equation, containing likewise the third power of the unknown quantity, belongs to cubic equations, and its resolution requires particular rules.

624 There are, therefore, only three kinds of terms in an equation of the second degree:

1. The terms in which the unknown quantity is not found at all, or which is composed only of known numbers.
2. The terms in which we find only the first power of the unknown quantity.
3. The terms which contain the square, or the second power, of the unknown quantity.

So that $$x$$ representing an unknown quantity, and the letters $$a$$, $$b$$, $$c$$, $$d$$, etc. the known quantities, the terms of the first kind will have the form $$a$$, the terms of the second kind will have the form $$bx$$, and the terms of the third kind will have the form $$cx^2$$.

625 We have already seen, how two or more terms of the same kind may be united together, and considered as a single term.

For example, we may consider the formula $$ax^2 - bx^2 + cx^2$$ as a single term, representing it thus, $$(a - b + c)x^2$$; since, in fact, $$(a - b + c)$$ is a known quantity.

And also, when such terms are found on both sides of the sign =, we have seen how they may be brought to one side, and then reduced to a single term. Let us take, for example, the equation,

$2x^2-3x+4=5x^2-8x+11;$

we first subtract $$2x^2$$, and there remains

$-3x+4=3x^2-8x+11;$

then adding $$8x$$, we obtain,

$5x+4=3x^2+11;$

lastly, subtracting 11, there remains $$3x^2 = 5x - 7$$.

626 We may also bring all the terms to one side of the sign =, so as to leave zero, or 0, on the other; but it must be remembered, that when terms are transposed from one side to the other, their signs must be changed.

Thus, the above equation will assume this form, $$3x^2-5x+7 = 0$$; and, for this reason also, the following general formula represents all equations of the second degree;

$ax^2 \pm bx \pm c = 0;$

in which the sign ± is read plus or minus, and indicates, that such terms as it stands before may be sometimes positive, and sometimes negative.

627 Whatever therefore be the original form of a quadratic equation, it may always be reduced to this formula of three terms. If we have, for example, the equation

$\frac{ax+b}{cx+d} = \frac{ex+f}{gx+h}$

we may, first, destroy the fractions; multiplying, for this purpose, by $$cx+d$$, which gives

$ax+b=\frac{cex^2+cfx+edx+fd}{gx+h},$

then by $$gx+h$$, we have

$agx^2+bgx+ahx+bh=cex^2+cfx+edx+fd,$

which is an equation of the second degree, reducible to the three following terms, which we shall transpose by arranging them in the usual manner

$\begin{Bmatrix} ag\\ -ce \end{Bmatrix} x^2 + \begin{Bmatrix} +bg\\ +ah\\ -cf\\ -ed \end{Bmatrix} x + \begin{Bmatrix} +bh\\ -fd \end{Bmatrix} = 0.$

We may exhibit this equation also in the following form, which is still more clear

$(ag-ce)x^2+(bg+ah-cf-ed)x+bh-fd=0.$

628 Equations of the second degree, in which all the three kinds of terms are found, are called complete, and the resolution of them is attended with greater difficulties; for which reason we shall first consider those, in which one of the terms is wanting.

Now, if the term $$x^2$$ were not found in the equation, it would not be a quadratic, but would belong to those of which we have already treated; and if the term, which contains only known numbers, were wanting, the equation would have this form, $$ax^2 \pm bx = 0$$, which being divisible by $$x$$, may be reduced to $$ax \pm b = 0$$, which is likewise a simple equation, and belongs not to the present class.

629 But when the middle term, which contains the first power of $$x$$, is wanting, the equation assumes this form, $$ax^2 \pm c = 0$$, or $$ax^2 = c$$; as the sign of $$c$$ may be either positive, or negative

We shall call such an equation a pure equation of the second degree, and the resolution of it is attended with no difficulty; for we have only to divide by $$a$$, which gives $$x^2 = \dfrac{c}{a}$$; and

$x=\surd \frac{c}{a};$

which means the equation is resolved.

630 But there are three cases to be considered here. In the first, when $$\frac{c}{a}$$ is a square number (of which we can therefore really assign the root) we obtain for the value of $$x$$ a rational number, which maybe either integral, or fractional. For example, the equation $$x^2=144$$, gives $$x = 12$$. And $$x^2=\frac{9}{16}$$, gives $$x=\frac{3}{4}$$.

The second case is, when $$\frac{c}{a}$$ is not a square, in which case we must therefore be contented with the sign $$\surd$$. If, for example, $$x^2 = 12$$, we have $$x =\surd 12$$, the value of which may be determined by approximation, as we have already shown.

The third case is that, in which $$\frac{c}{a}$$ becomes a negative number: the value of $$x$$ is then altogether impossible and imaginary; and this result proves that the question, which leads to such an equation, is in itself impossible.

631 We shall also observe, before proceeding farther, that whenever it is required to extract the square root of a number, that root, as we have already remarked, has always two values, the one positive and the other negative. Suppose, for example, we have the equation $$x^2 = 49$$, the value of $$x$$ will be not only +7, but also -7, which is expressed by $$x= \pm 7$$. So that all those questions admit of a double answer; but it will be easily perceived that in several cases, as those which relate to a certain number of men, the negative value cannot exist.

632 In such equations, also, as $$ax^2= bx$$, where the known quantity $$c$$ is wanting, there may be two values of $$x$$, though we find only one if we divide by $$x$$. In the equation $$x^2=3x$$, for example, in which it is required to assign such a value of $$x$$, that $$x^2$$ may become equal to $$3x$$, this is done by supposing $$x = 3$$, a value which is found by dividing the equation by $$x$$; but, beside this value, there is also another, which is equally satisfactory, namely, $$x=0$$; for then $$x^2=0$$, and $$3x= 0$$. Equations therefore of the second degree, in general, admit of two solutions, whilst simple equations admit only of one.

We shall now illustrate what we have said with regard to pure equations of the second degree by some examples.

633 Question 1. Required a number, the half of which multiplied by the third, may produce 24.

Let this number be $$x$$; then by the question $$\frac{1}{2}x$$, multiplied by $$\frac{1}{3}x$$, must give 24; we shall therefore have the equation $$\frac{1}{6}x^2 = 24$$.

Multiplying by 6, we have $$x^2 = 144$$; and the extraction of the root gives $$x = \pm 12$$. We put ±; for if $$x=+12$$, we have $$\frac{1}{2}x = 6$$, and $$\frac{1}{3}x = 4$$: now, the product of these two numbers is 24; and if $$x = - 12$$, we have $$\frac{1}{2}x = -6$$, and $$\frac{1}{3}x = -4$$, the product of which is likewise 24.

634 Question 2. Required a number such, that being increased by 5, and diminished by 5, the product of the sum by the difference may be 96.

Let this number be $$x$$, then $$x+5$$, multiplied by $$x-5$$, must give 96; whence results the equation,

$x^2-25=96.$

Adding 25, we have $$x^2= 121$$; and extracting the root, we have $$x=11$$. Thus $$x + 5 = 16$$, also $$x - 5 = 6$$; and, lastly, 6 ⋅ 16 = 98.

635 Question 3. Required a number such, that by adding it to 10, and subtracting it from 10, the sum, multiplied by the difference, will give 51.

Let $$x$$ be this number; then $$10+x$$, multiplied by $$10-x$$, must make 51, so that $$100 - x^2 = 51$$. Adding $$x^2$$, and subtracting 51, we have $$x^2 = 49$$, the square root of which gives $$x = 7$$.

636 Question 4. Three persons, who had been playing, leave off; the first, with as many times 1 crowns, as the second has three crowns; and the second, with as many times 17 crowns, as the third has 5 crowns. Farther, if we multiply the money of the first by the money of the second, and the money of the second by the money of the third, and, lastly, the money of the third by that of the first, the sum of these three products will be 3830⅔. How much money has each?

Suppose that the first player has $$x$$ crowns; and since he has as many times 7 crowns, as the second has 3 crowns, we know that his money is to that of the second, in the ratio of 7:3.

We shall therefore have 7:3 ∷ $$x$$:$$\frac{3}{7}x$$, the money of the second player.

Also, as the money of the second player is to that of the third in the ratio of 17:5, we shall have 17:5 ∷ $$\frac{3}{7}x$$:$$\frac{15}{119}x$$, the money of the third player.

Multiplying $$x$$, or the money of the first player, by $$\frac{3}{7}x$$, the money of the second, we have the product $$\frac{3}{7}x^2$$: then, $$\frac{3}{7}x$$, the money of the second, multiplied by the money of the third, or by $$\frac{15}{119}x$$, gives $$\frac{45}{833}x^2$$; and, lastly, the money of the third, or $$\frac{15}{119}x$$, multiplied by $$x$$, or the money of the first, gives $$\frac{15}{119}x^2$$. Now, the sum of these three products is

$\frac{3}{7}x^2+\frac{45}{833}x^2+\frac{15}{119}x^2;$

and reducing these fractions to the same denominator, we find their sum $$\frac{507}{833}x^2$$, which must be equal to the number 3830⅔.

We have therefore, $$\frac{507}{833}x^2 = 3830\frac{2}{3}$$.

So that $$\frac{1521}{833}x^2 = 11492$$, and $$1521x^2$$ being equal to 9572836, dividing by 1521, we have $$x^2 = \frac{9572836}{1521}$$; and taking its root, we find $$x = \frac{3094}{39}$$. This fraction is reducible to lower terms, if we divide by 13; so that $$x = \frac{283}{3} = 79 \frac{2}{3}$$; and hence we conclude, that $$\frac{3}{7}x = 34$$, and $$\frac{15}{119}x=10$$.

The first player therefore has 79⅓ crowns, the second has 34 crowns, and the third 10 crowns.

Remark. This calculation may be performed in an easier manner; namely, by taking the factors of the numbers which present themselves, and attending chiefly to the squares of those factors.

It is evident, that 507 = 3 ⋅ 169, and that 169 is the square of 13; then, that 833 = 7 ⋅ 119, and 119 = 7 ⋅ 17: therefore

$\frac{3 \cdot 169}{17 \cdot 49}x^2=3830\frac{2}{3},$

and if we multiply by 3, we have $$\frac{9 \cdot 169}{17 \cdot 49}x^2=11492$$. Let us resolve this number also into its factors; and we readily perceive, that the first is 4, that is to say, that 11492 = 4 ⋅ 2873; farther, 2873 is divisible bv 17, so that 2873 = 17 ⋅ 169. Consequently, our equation will assume the following form,

$\frac{9 \cdot 169}{17 \cdot 49}x^2=4\cdot 17\cdot 169,$

which, divided by 169, is reduced to

$\frac{9}{17 \cdot 49}x^2=4\cdot 17;$

multiplying also by 17 ⋅ 49, and dividing by 9, we have

$x^2=\frac{4 \cdot 289 \cdot 49}{9},$

in which all the factors are squares; whence we have, without any further calculation, the root

$x = \frac{2 \cdot 17\cdot 7}{3}=\frac{238}{3}=79\frac{1}{3},$

as before.

637 Question 5. A company of merchants appoint a factor at Archangel. Each of them contributes for the trade, which they have in view, ten times as many crowns as there are partners; and the profit of the factor is fixed at twice as many crowns, per cent, as there are partners. Also, if ¹⁄₁₀₀ part of his total gain be multiplied by 2²⁄₉, it will give the number of partners. That number is required.

Let it be $$x$$; and since, each partner has contributed $$10x$$, the whole capital is $$10x^2$$. Now, for every hundred crowns, the factor gains $$2x$$, so that with the capital of $$10x^2$$ his profit will be $$\frac{1}{5}x^3$$. The ¹⁄₁₀₀ part of his gain is $$\frac{1}{500}x^3$$; multiplying by 2²⁄₉, or by ²⁰⁄₉, we have $$\frac{20}{4500}x^3$$, or $$\frac{1}{225}x^3$$, and this must be equal to the number of partners, or $$x$$.

We have, therefore, the equation $$\frac{1}{225}x^3 = x$$, or $$x^3=225x$$; which appears, at first, to be of the third degree; but as we may divide by $$x$$, it is reduced to the quadratic $$x^2=225$$; whence $$x = 15$$.

So that there are fifteen partners, and each contributed 150 crowns.

Editions

1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.