« ΠροηγούμενηΣυνέχεια »
· Let us take the general equation.
2 + px=9. This, we have seen, may represent any equation whatever of the second degree, p and q being any known quantities and either positive or negative. If p = 0 the equation becomes
which is a pure equation or an equation with two terms.
If we make the first member of the equation x2 + px=%, a complete second power, by the above rules, it becomes
The first member of this equation is the difference of two second powers, which, Ait. XIII, is the same as the product of the sum and difference of the numbers.
The sum is x + 9 + m, and the difference is x + £ —m, and their product is
(x + - m) (x + .+ m) = 0.
In this equation, the first member consists of two factors, and the second is zero. Now the first member of the above Equation will be equal to zero, if either of its factors is equal to zero. For if any number be multiplied by zero, the product is zero.
Making the first factor equal to zero,
*=- + m. Making + +m=0.. gives
=- -m. Either of these values of x must answer the conditions of the equation.
N. B. Though either value answers the conditions separately, they cannot be introduced together, for being different, their product cannot be x*.
Instead of m put its value, and the values of x become
which are the values we had obtained above. stration is essentially that of M. Bourdon.)
Let us take again the general equation. "... =- (2+)*
Since the expression contains a radical quantity, that is, a quantity of which the root is to be found, in order to be able to find the value of it, we must be able to find the root either exactly or by approximation. Now there is one case in which it is impossible to find the root. It is when q is negative and greater than In which case the expression q + is negative ; and it has been shown above, that it is impossible to find the root of a negative quantity. In all other cases the value of the equation may be found.
In all cases if q is positive, the first value will be positive, and answer directly to the conditions of the question proposed. For the radical (2 + ) is necessarily greater than , be
cause the roc
? - ; therefore the expression
necessarily of the same sign as the
The second value is for the same reason essentially negative,
for both Land (2 + 4) are negative. This value, though
it fulfils the conditions of the equation, does not answer the conditions of the question, from which the equation was derived ; but it belongs to an analogous question, in which the x must be put in with the sign — instead of t; thus x? — pr=9, which
a value, which differs from the
first only by the sign before
If q is actually negative, the equation becomes
x+px = -9, and the values are
In order that it may be possible to find the root, q must be less than . When this is the case, the two values are real. Since (2-a)' is smaller than , it follows that both values are negative if p is positive in the equation ; that is, if 20 +px=-9, which gives
x=-=(-2)*;. and both positive if p is negative in the equation, that is, x' px=-9, which gives
* = = (4-2)* When both values are negative, neither of them answers directly to the conditions of the question; but if — x be put into the original equation instead of x, the new equation will show what alteration is to be made in the cnunciation of the question; and the same values will be found for x as before, with the exception of the signs.
If in this equation q is greater than in the quantity (4 -9)* becomes negative, and the extraction of the root cannot be performed. The values are then said to be imoginary. .
1. It is required to find two numbers whose sum is p, and whose product is q.
Let x = one of the numbers, then p- x = the other.
px - x = 1; Changing signs que — px=-9.
This.example presents the case above mentioned, in which p and q are both negative.
The values are 9 and 6, both positive, and both answer the conditions of the question. And these are the two numbers required, for 9 + 6 = 15,9 x 6 = 54. This ought to be so, for x in the equation represents either of the numbers indifferently. Indeed whichsoever x be put for, p- & will represent the other; and px - x? will be their product. Again let p = 16 and q = 72.
x = + ( 256 — 72)* = 8 +(-3)*. Here ( 8) is an imaginary quantity, therefore both values are imaginary. ,
In order to discover why we obtain this imaginary result, let us first find into what two parts a number must be divided, that the product of the two parts may be the greatest possible quantity.
In the above example, p represents the sum of the two numbers or parts, let d represent their difference, then
+ = the greater, and . - = the less. Ar. IX. Their product is
(+ -) (% - - ) = Art. XIII. The expression is evidently les than e' so long as .
dis greater than zero; but when d=0, the expression becomes