In fact, decompose 72, into 64+8; we have √64-8, √8= 2+ a fraction; whence √64+ √8=10+ a fraction; again 72 =49+23; we have √49=7, √23=4+ a fraction; hence √49+ √23=11+ a fraction.

For a third example we will consider the expression m2x2 + n2

which it is required to render a minimum, (m be

Now, in order that the two values of x, corresponding to a value of y may be real, it is necessary that (m2-n2)2y2, should not be less than 4m2n2, and consequently that y should not be

cal will disappear, and the value of x becomes

110. These examples will suffice to point out the course to be followed, in the solution of questions of this kind. After having formed the algebraic expression of the quantity which is susceptible of becoming a maximum or minimum, place it equal to some letter, as y. If the equation thus obtained is of the second degree in x (x denoting the variable quantity which enters into the expression), resolve it with reference to x, then place the quantity under the radical equal to zero, and from this last equation find the value of y, which will represent the required maximum or minimum. Substitute this value of y in the expression for x, and we will have the value of this variable, which will satisfy the enunciation.

N. B. If it should happen that the quantity under the radical remains essentially positive, whatever value be given to y, we may conclude that the proposed expression can pass through every possible state of magnitude; in other words, it would have infinity for its maximum, and zero for the minimum.

For another example take the expression

4x+4x-3

6(2x+1)

find whether it is susceptible of a maximum or minimum.

; and

3y-1 1

6(2x+1)

4x2-4(8y-1)x=6y+3, whence x= 2 2

Now whatever value is given to y, the quantity under the radical will always be positive. Therefore y, or the proposed expression, can pass through every state of magnitude.

In the preceding examples, the quantity under the radical, in the value of x, contained but two parts, one affected with y or y2, and the other entirely known; and the maximum or minimum of which the function was susceptible was easily obtained. But it may happen that the quantity under the radical is a trinomial of the second degree of the form my2+ny+p. In this case the question becomes more difficult, and to be able to resolve it, it is necessary to demonstrate some properties relative to these trinomials.

Properties of Trinomials of the Second Degree.

111. Every algebraic expression which can be reduced to the form my3 +ny+p, is called a trinomial of the second degree; m, n, and p, being known quantities,+ or -, y designating a variable, that is, a quantity which may pass through different states of magnitude.

Thus,

3y2-5y+7, 9y2+2y+5,

(a−b+2c)y2+4b2y-2ac2 +3a2b,

are called trinomials of the second degree.

Placing the trinomial my2+ny+p equal to zero, we have

n 1

my2+ny+p=0, whence y= + √ n2-4mp.

2m 2m

Three principal hypotheses can be made with respect to the

nature of the values of y. We may have n2-4mp>0, or positive; in which case, the two roots are real and unequal,

+ or -.

Or we may have n2—4mp=0; in which case the two roots are real and equal.

Or, lastly, we may have n3—4mp<0, or negative; then the two roots are imaginary.

Therefore the following properties will result from these different cases:

1st. When a trinomial of the second degree is such, that by placing it equal to 0, and resolving the resulting equation, we obtain two real and unequal roots, every quantity (positive or negative) comprehended between the two roots, and substituted for y in the trinomial, will necessarily give a result, with a sign contrary to that affecting the coefficient of y2; but every quantity not comprised between the two roots, and substituted for y, will give a result with the same sign as that of the coefficient of y2.

For, let y' and y' denote the two roots (supposed real) of the

equation my2+ny+p=0, or m

n

(

y2 +

The first member my2+y+

(

m

m

-

n Ρ

m

-0.

y + 2) = 0.

m

1)=

can be put under the

form (No. 98) m(y—y') (y—y'). Therefore we have the identity

my2+ny+p=m(y—y') (y—y').

Let a be a quantity comprised between y' and y', that is, such that a> or <y', but a< or >y"; there will result a-y'> or <0, but a—y'<or>0; hence the two factors a—y', a—y" have contrary signs, therefore their product is negative. Consequently, m (a-y') (a-y'), or its value, ma2+na+p, has a sign contrary to that of m.

If, on the contrary, we suppose at the same time a> or <y', and a> or <y", whence a-y'> or <0, and a-y"> or <0, the two factors will have the same sign; hence their product, . . . . (a-y') (a-y''), is positive, and consequently m(a-y') (a—y''), or ma2+na+p, is of the same sign as m.

....

2d. When the two roots are real and equal, every quantity different from that which reduces the trinomial to 0, will, when

substituted in the trinomial, give a result of the same sign as the

coefficient of y2.

For since the two roots are equal, we have the relation.

2

quantity (y+) will be positive for every value substituted

3d. When the two roots are imaginary, every real positive or negative quantity, substituted in the place of y, will give a result of the same sign as that of the coefficient of y2.

For since the two roots are imaginary, we have the relation n2-4mp <0, whence 4mp>n; or (105) dividing by 4m2.... P na > 4m2

m

Take P

positive. There will result

n

k denoting a quantity essentially

Ρ

Im 4m2

my3 + ny+p, or m (y2 + " y + 2) = m (y2+ y + + k2)

(3

n

m

n2

4m2

2

2m

which always has the same sign as m, whatever value may be substituted for y.

112. The second case naturally leads us to speak of a proposition of frequent use in analysis.

When a trinomial of the second degree, my2+ny+p, is a perfect square, there is between the coefficients the relation n2-4mp=0.

For if this trinomial is a perfect square, and of the form (m'y+n'), by placing it equal to 0, the two roots of the resulting equation will be equal. Now in order that they may be equal, the quantity under the radical, or n2-4mp, must be nothing. Therefore we have the relation n2-4mp=0.

Reciprocally. When there is, between the coefficients the relation n2-4mp-0, the trinomial is a perfect square; for

113. The following examples will show the use of these properties in the solution of questions of maximum and minimum.

Let it be proposed to determine whether, in making a vary, x2-2x+21

6x-14

the function

may pass through every state of mag

=y; whence

x2-2 (3y+1)x=-21-14y.

From which we have x=3y+1±√9y2-8y-20. In order that x may be real, it is necessary that . . 9y2-8y-20 should be positive. Now, placing this equal to 0, we have

y2

9 20

0; whence y=2 and y=

10
9

These two values of y being real, it follows, from the first of the above properties, that by giving y values comprised between

10
9

2 and -, such as 1, 0, -1, ... the value of the trinomial will

be negative, since the coefficient of y2 is positive; but by giving

to y values which are not comprised between 2 and -

10

as

9

3, 4...., or -2, -3, -4,.,...., the result of the substitu—2, —4, tion will be positive. Hence we perceive that 2 is in absolute numbers, the minimum value that y should have, in order that x may be real. If, in the above expression we make y=2, the radical will disappear, and we find x=7.