Substituting the first, in which the radical has the sign +,
(p+ √ Ap2-q)2+p(−{p+√{p2−q)+q=0,

2

2

or p2-p√ p2-q+{p2—q¬} p2+p√}p2¬q+q=0; Reducing the like terms, the first member of the equation becomes 0, and consequently the equation is satisfied by the value

x = − } p + √ { p2-q.

The second value of x is verified in the same manner, and with no change except that of the sign for + before the radical

120. In the equation x2+px+q=0, p is the coefficient of x and q the term wholly known, transposed to the first member; the solution, x=−!p± p-q, may therefore be expressed in words as follows:

When an equation of the second degree is reduced to the form x2+p+q=0, the unknown quantity is equal to half the coefficient of x taken with a contrary sign, plus or minus the square root of the sum formed by adding to the square of this half the term which is wholly known, this term being also taken with a contrary sign.

To apply this rule to examples:

1st. Let x2-10x+9=0; required the values of x?

This example has the form required by the rule. The coefficient of x is -10; therefore half the coefficient of x taken with a contrary sign is +5; the term wholly known is 9.

Therefore the rule gives this result,

x=5±√25—9=5+√16=5+4;

And the two values of rare consequently x=5+4=9,

and x=5-4=1.

; required the values of x?

To reduce this equation to the form x2+px+q=0, it is necessary to remove the denominators; thus, 3x2+18=2x+26; next, to transpose the terms to the left member;

whence 3r2-2x+18-26=0,

or 3x2-2x-8=0;

and lastly, to divide all the terms by 3; making the division,

x2-3x-3=0.

Whence p+ 2 = + +

In this example the two values of x are equal.

4th. Given 100x2-100x+41=0; required the values of x? Dividing the terms by 100, x2-x+11=0.

The values of x in this example are imaginary. The expression √

is reduced to -1, by Article 104.

16

352

121. In a preceding article (Art. 103) the first member of the equation, -A and x+√A. A like -A=0, is decomposed into the two factors xdecomposition can be made in the general equation,

x2+px+q=0.

Replacing 2+px by the equivalent expression (r+p)2-p2, the result is, x2+px+q=(x+}p)2 —(†p2−q); and since 4p2-q can be changed into (Ap-q), the difference (x+p)2-√(tp2-q), becomes

(x+} p−√ } p2−q)× (x+3p+√\p2—q),

which corresponds with the result obtained in the case x2-A=0.

Since the values of r ought to render the last product equal to zero, this decomposition furnishes a new process for the resolution of equation 2; for Thereit is evident that a product is equal to zero if either of its factors is equal to zero, but that it cannot be zero if neither of its factors is zero. fore the values of x which satisfy equation 2, ought to be given by the equations

A.

These equations give the roots already obtained, viz.

x=−{p+√}p2-q

B.

From these results it is concluded, that the two factors of the first member of the equation x2+px+q=0 are the differences which are obtained by subtracting the roots severally from x, so that, expressing the two roots by the characters x', x' respectively,

x2+px+q=(x-x')(x-x').

If the multiplication indicated in the second member of the last equation is performed, +px+q=x2−(x+x')x+x'x'; and since the two members

of this equation must be equal, term by term, it follows that

p=—x'—x''
q=x'x'.

These relations, which can be verified directly by means of the values (Equation 4) are enunciated in words thus:

In every equation of the second degree reduced to the form x2+px+9=0, the coefficient p of the second term is equal to the sum of the two roots, taken with contrary signs; and the term entirely known, q, is equal to the product of the roots.

When it is known that the two roots of an equation of the second degree are real, the preceding relations immediately make known the nature of the roots. For example, admitting that the roots of the equation x2-2x-7=0 are real, it is inferred that they are of different signs, for their product is equal to -7, the term which is entirely known; also, that the greater is positive, for their sum is equal to +2, the coefficient of x taken with a contrary sign.

122. The general equation of the second degree being x2+px+q=0, the general values of the roots are

2

p ✓ -q, in which the term p2, being

4

a square, is positive, whatever be the sign of p.

a. If q is negative in the general equation, the quantity p2-q, or, in this case, p2+9, is positive and greater than p2; therefore p2+q is a real quantity, and greater in absolute value than p; therefore -p placed before the radical does not change the signs of the two values of the radical; therefore the two values of x have contrary signs.

It is, besides, evident that the greater is of the same sign as-p, and consequently of the sign contrary to p.

b. If q=0, the two values of z are,

x=−}p+ / £ 2=−{p+}p=0,

In this case the general equation is reduced to x2+px=0, or x(x+p)=0, and it is evident that its roots are x=0 and x=—-p.

c. If q is positive and less than p2, the radical √p2-q is real, but less than p; therefore the two values of x are of the same sign as the term -p placed before the radical, that is to say, they are both positive when p is negative, and both negative when p is positive.

d. If q=p, the two values of x are reduced to one, viz., x—— -p. In this case, since q=p2,

x2+px+p2=0, or (x+p)2=0, or (x+1p)(x+p)=0; the last of which equations, it is evident, can be verified by the value x=—ip, and by no other.

e. If q is positive, and greater than p, the quantity p2-q under the radical sign is negative, and the two values of x are imaginary.

Changing the signs of the quantity p2-q, and making the square root of q-p2, p2-q=—p2;

Therefore

p2¬q=±√−r2=+r√−1,

and r=- -{{p+r√−1.

From p2-q=—r2 is obtained q={p2+r2.

Whence the general equation becomes

x2+px+p2+r22=0, or (x+3p)2+r2=0.

The first member of this equation is the sum of two squares. Under this form the reason wherefore the two values of r are imaginary is obvious; namely, that there can exist no value, whether positive or negative, which being substituted for x can render the sum of these two squares equal to

zero.

123. To resolve, in the ordinary manner, the equation

ax2+ bx+c=0,

all the terms are divided by a, the coefficient of x;

In this expression, according as b2-4ac>0, or =0, or <0, the values of x are real and unequal, real and equal, or imaginary.

a. Suppose that a, the coefficient of 2, is diminished until it becomes equal to zero, the two values of x, taken separately, are

If a were a common factor of the numerator and denominator of this expression, by suppressing the common factor, and then making a=0, the true value of x would be obtained; this, in the present instance, cannot be accomplished, but the difficulty can be avoided. Multiplying the terms of the fractional value of x by -b−√/b2-4ac, the result is

The numerator being now the product of the sum and difference of the quantities -b and b-4ac, it is equal to the difference of the squares of these quantities; that is, to

b2-(b2-4ac), or to 4ac;

2a, which is a common factor of both terms of this value of x, being

с

and therefore is the true value of the root which at first appeared

under the form

-26
0'

With respect to the value, x= it is to be observed that the divisor, O, may be considered in this case as the limit of decreasing magnitudes, whether positive or negative; and that, for this reason, the infinite value ought to have the ambiguous sign +. Whence, the values of x, deduced from the equation ar2+br+c=0 in the hypotheses a=0, are

It deserves notice that in this particular case x has three values while in the general case it has only two.

To show that these values satisfy the equation ax2+bx+c=0 when a=0,

let the equation be reduced to the form

-bx-c
x2

=0.

The problem consequently is, when a=0, to find values which render the fractional expression

-bx-c
x2

=0.

с

bc

Now it is evident that the value x=- renders the numerator +-c=0,

b

that these expressions become zero, for x=+ since

b. Let a=0, b=0.

In this particular case the two general values of r become g.

It has been already shown that the first general value of x can be changed

Transforming the second value of x in the same manner it becomes first (—b—√/b2 —4ac) (−b+√b2−4ac)

Making a=0, b=0, both the values of x, thus transformed, give x=x; and in this case (as in the preceding) the symbol ought to have the sign prefixed to it.

124. An equation which contains only two powers of the unknown quantity, the exponent of the one power being double that of the other, may be put under the form

x2+px+9=0

1.

Considering as the unknown quantity, a2m is the square of the unknown quantity; consequently the equation can be resolved like an equation of the 2d degree.

2

Whence --}p±√¿p2—q.

Denoting-p+ p2—q by a, and −1p-√4p2—q by b, the two values of 2 are ra, x"=b.

When these equations can be resolved it is evident that all the values of x are known.

The only case which shall be considered at present is that of m=2, Equation 1, in this case, becomes

x++px2+q=0

Making 2 at first the unknown quantity,

2.

x2 = p + √ } p2 -q, x2=-}p-√ } p2-q.

Since each of the values of x2 gives two values of x, equation 4 has four solutions; viz.

x=+ √3p+ √ p2—q, x=±√{p—√{p2 —q. Assuming the particular instance, a-12x264-0, it is necessary, in the preceding formula, to replace p by -12 and q by -64; then

x=+6+√36+64

+6+100=+√6+10=√16=+4,

x=+√6−√36+64=+√‍6−√100=±√/6−10=+√−4=2√—1.

Whence the proposed equation has four roots; two real, which are x=+4, x=-4, and two imaginary, x=+2√−1,x=· :—2√—1.

125. The principles by means of which equations of the first degree containing two or more unknown quantities are resolved receive their application in the resolution of equations of the second degree also. Example. Let x+y=a, x2+y2=b2; required the values of x, y From the 1st equation, y=a-x.. y2= (a—x)2=a2—2ax+x2. Substituting this value, instead of y, in the 2d equation,

x2+a2-2ax+x2=b2,
or, 2x2-2ax-b2—a2 ;

?

Substituting this value of x in the equation y=a-x, the values of y are,