CHAPTER II

SOLUTION OF EQUATIONS OF THE SECOND DEGREE

403. Solution of the equation ax = b.-Every equation in which, besides given constants, the first power of the unknown quantity, x, alone occurs, may by multiplication, addition, and subtraction, be reduced to the form

Division is the last step involved in finding the value of x. In solving an equation of the first degree in x, there are involved only the four fundamental operations of common Algebra.

404. The Solution of the Pure Quadratic Equation ax = b. An equation which involves a2 only, in addition to given constants, may be reduced by the four fundamental operations of common Algebra, as has been explained above, to the forms,

In order to find the values of x, it is necessary to employ a fifth operation, the extraction of the square root (Theorems II and III, 2400, 401), which gives

x=

There will be two values of x, namely,

because

and

x1 = + √ A2 and x = -VA;

(+ √A)2 = A = x2

(— √ Ã)2 = A = x2.

A,

Both values of x will be real when A is positive; but if A is negative, the equation x2-4, can not be satisfied by any real values of x, since x2 must always be positive, and can not be equal

These quantities, x, and x, are called imaginary values of x.

whence

(3) x2—(VÃ)2=(x+VÃ)(x−√Ā)= 0. [894]

Equation (3) can be satisfied by placing each factor equal to zero.

Hence the equation may be written in the form,

20. (a + bx)+(ax—b)2 = 2 (a2x2 + b2).

21. (7x) (9 — x) + (7 — x) (9 + x) = 76.

22. (2x+7) (5x — 9) + (2x-7) (5x+9)= 1874.

23. (1 + x) (2 + x) (3 + x) + (1 − x) (2 − x) (3 — x) = 120.

24. (2x+3) (3x + 4) (4x+5) — (2x-3) (3x-4) (4x-5) 184. 25. (x+a+b) (x − a + b)

+ (x + a − b ) (x — a — b) = 0.

26. (a + bx) (b − ax) + (b + cx) (c − bx) + (c + ax) (a — cx) = 0. 27. (a+x) (x) + (1 + ax) (1 − bx) = (a + b) (1 + x2). 28. (a + 5b + x) (5a + b + x) = 3 (a + b + x)2. 29. (9a7b+3x) (9b7a+3x) = (3a+3b+ x)2.

THE SOLUTION OF THE EQUATION ax + bx + c = 0

405. An equation which involves the first and the second powers only of the unknown quantity x, i. e., x and x2, besides given constants, may be reduced by the four fundamental operations of Algebra to the form

In the case of these two equations the problem set is to find what and how many values of x there are involving a, b, and c, or p and q which will satisfy equation (1) or (2). It is to be remarked that the equality (=) does not exist for all values of x, but for two only (as will be proved).

It will be found that the final solutions of equations (1) and (2) are obtained by using only the operations employed in solving the simple equation of one unknown quantity, which can be reduced to the form ax = b, and the pure quadratic equation, which can be reduced to the form x2 = 4; i. e., by using the four fundamental operations and the extraction of the square root.

The method for solving equation (1), which is about to be explained, will be illustrated by an example. Consider the equation

9x+4x= 13.

The coefficient of x is the square of 3 and the equation may be solved by adding to both members such a quantity as will make the first member a trinomial square. By Theorem V, 402,

(y + z)2 = y2+ 2 yz + z2;

from which it follows that the third term of a trinomial square is

Hence, the third term of a trinomial, which is a perfect square, is the square of the quotient of the middle term by twice the square root of the first term. Therefore, in this example,