236. Solve

x2-6x+13=0.

By transposition, x2-6x=-13;

add 32,

x2-6x+32=-13+9= −4.

If we try to extract the square root we have

But 4 can have no square root, exact or approximate, because any number, whether positive or negative, if multiplied by itself, gives a positive result. In this case the quadratic equation has no real root; and this is sometimes expressed by saying that the roots are imaginary or impossible.

Here we first clear of fractions by multiplying by 4(x2-1), which is the least common multiple of the denominators.

Multiply by 570, which is the least common multiple of 15 and 190; thus

Clear of fractions; thus

(x+3)(x-2)(x-1)+(x-3)(x+2) (x − 1)

=(2x-3)(x+2) (x − 2);

that is, a-7x+6+x3-2x2-5x+6=2x3-3x2 — 8x+12;

that is,

2x3 — 2x2-12x+12=2x3-3x2-8x+12;

We have given the last three lines in order to complete the solution of the equation in the same manner as in the former examples; but the results may be obtained more simply. For the equation x2-4x=0 may be written (x-4)=0; and in this form it is sufficiently obvious that we must have either x-4=0, or x=0, that is, x=4 or 0.

The student will observe that in this example 23 is found on both sides of the equation, after we have cleared of fractions; accordingly it can be removed by subtraction, and so the equation remains a quadratic equation.

240. Every quadratic equation can be put in the form x2+px+q=0, where p and q represent some known numbers, whole or fractional, positive or negative.

For a quadratic equation, by definition, contains no power of the unknown quantity higher than the second. Let all the terms be brought to one side, and, if necessary, change the signs of all the terms so that the coefficient of the square of the unknown quantity may be a positive number; then divide every term by this coefficient, and the equation takes the assigned form.

For example, suppose 7x-4x=5. Here we have

242. We have thus obtained a general formula for the roots of the quadratic equation x2+px+q=0, namely, that a must be equal to

−p+ √(p2-4q)

2

or to P√(p2-4q)

2

We shall now deduce from this general formula some very important inferences, which will hold for any quadratic equation, by Art. 240.

243. A quadratic equation cannot have more than

two roots.

For we have seen that the root must be one or the other of two assigned expressions.

244. In a quadratic equation where the terms are all on one side, and the coefficient of the square of the unknown quantity is unity, the sum of the roots is equal to the coefficient of the second term with its sign changed, and the product of the roots is equal to the last term.

For let the equation be x2+px+q=0;

the sum of the roots is

−p+ √(p2−4q) + −p− √(p2-4q), that is —p;

2

the product of the roots is

that is

2

245. The preceding Article deserves special attention, for it furnishes a very good example both of the nature of the general results of Algebra, and of the methods by which these general results are obtained. The student should verify these results in the case of the quadratic equations already solved. Take, for example, that in Art. 232; the equation may be put in the form

247. The general formulæ given in Arts. 241 and 246 may be employed in solving any quadratic equation. Take for example the equation 3x2-4x-55=0; divide by 3, thus we have

Take the formula in Art. 241, which gives the roots of

4 3'

x2+px+q=0; and put p= and q=

55

thus obtain the roots of the proposed equation.

3

; we shall

But it is more convenient to use the formula in Art. 246, as we thus avoid fractions. The proposed equation being 3x2-4x-55=0, we must put a=3, b=-4, and c=— in the formula which gives the roots of ax2+bx+c=0,

-b+ √(b2-4ac)

-55,

that is, in

Thus we have

that is,

6

6