401. THEOREM III.-To square a fraction, take the quotient of the square of the numerator by the square of the denominator.

THEOREM IV.-Conversely, to extract the square root of a fraction, take the quotient of the square root of the numerator by the square root of the denominator. Thus,

because the square of the second member, by Theorem III, is

402. THEOREM V.-The square of the sum of two quantities is the square of the first plus twice the product of the first by the second plus the square of the second.

By multiplying a + b by itself it is found that

(a+b)2= (a + b) (a + b) = a2+ 2 ab + b2.

Thus the squares of the binomials 2 x + 4 and 3 x

5 are

(2x+4)=4x2 + 2 2 x 4+16=4x2+16x+16,

(3x-5)2=[3x+(-5)]=9x2+2·3.x(-5)+(-5)2=9x2-30x+25.

Similarly,

(2 ax + b)2 = 4 a2x2 +4 abx+ b2,

( x + 2)2 = x2 + px + 22.

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 ar2 = b. An equation which involves a 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

Both values of x will be real when A is positive; but if A is negative, the equation x2 = — A, can not be satisfied by any real values of x, since a 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+√Ã)(x−√Ā)= 0. [394]

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

Hence the equation may be written in the form,

20. (a + bx)2 + (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. (ax) (b − x) + (1 + a.x) (1 — bx) = (a + b) (1 + x3).

28. (a+5b+x) (5a + b + x) = 3 (a + b + x)2.

29. (9a7b+ 3.x) (9b7a + 3x) = (3a + 3b + x)2.