THEORY OF QUADRATICS.

302. Having treated of the practical methods of solving quadratic equations, we will now proceed to consider certain general principles relating to quadratics.

303. Let us resume the general equation,

If we solve this equation, and represent one root by r and the other by r', we shall have

By adding these equations, and also multiplying them together, we

obtain

r+r' =−2a,
rr' == -b.

(3)

(4)

That is,

1.-The sum of the two roots is equal to the coefficient of x taken with the contrary sign.

2.-The product of the two roots is equal to the absolute term taken with the contrary sign.

304. From equations (3) and (4) in the last article, we have

2a=(rr'), and b = —rr”.

Substituting these values in (4), and transposing the absolute term, we have

or by factoring,

x2—(r+r')x+rr' = 0;

(x—r)(x—r') = 0.

Hence,

If all the terms of a quadratic equation be transposed to the first member, the result will consist of two binomial factors, formed by annexing the two roots with their opposite signs to the unknown quantity.

305. A Quadratic Expression is one which contains the first and second powers of some letter or quantity.

By the principle established in the preceding article, any quadratic expression whatever may be resolved into simple factors.

1. Let it be required to resolve the expression, x2+12x—45, in

to simple factors.

Assume

x+12x-450.

x= = 3, x = -15.

This equation readily gives

Hence,

x+12x-45(x-3) (x+15), Ans.

2. Separate 5-8x+3 into simple factors

We first separate the factor 5; thus,

We

may now factor the quantity within the parenthesis, as in the last example; thus,

8x

+

8x
5

5 25 25'

3

And the given quantity is factored as follows:
5x1-8x+3=5(x-1) (x-8), Ans.

EXAMPLES.

1. Resolve x2+2x-120 into simple factors.

Ans. (x-10) (x+12.)

2. Resolve x2-9x+14 into simple factors.

3. Resolve x2+8x+15 into simple factors.

Ans. (x-2) (x-7)

Ans. (x+3)(x+5).

4. Resolve x-35x+300 into simple factors.