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Ex. 5. Given a x + by + czd, a' x + b' y + c'z = d' a" x + b" y + c" z = d" to find x, y and z.

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a c'b'+ca' b'. ·ba'c'+bc' a" C b' a"

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a d'c"-a c'd"+ca' d"-da'c"+dc' a"-c d' a". a b' c'-a c' b"+c a' b". - b a'c'+b c' a' '—c b'a'

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a b' d' -a d' b'+d a' b"-ba'd'+b d' a".
a b' c'-a c' b"+c a' b" -b a' c''+bc' a"-

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-d b' a"

·c b' a'"

to find x, y, and z.

Ans. x =

2, y = 3, z = 4.

60, and (x + y)2 = 23, to find

Ans. x=10, y

Ex. 7. Given (x + y)



x and y.

= 2.

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A general algebraic problem is that in which all the quantities concerned, both known and unknown, are expressed by letters, or other general characters. Not only such problems as have their conditions proposed in general terms are here implied; every particular numerical problem may be made general, by substituting letters for the known quantities concerned in it; when this is done, the problem which was originally proposed in a particular form becomes general.

In solving a problem algebraically some letter of the alphabet must be substituted for an unknown quantity. And if there be more unknown quantities than one, the second, third, &c. must either be expressed by means of their dependence upon the first and one or other of the data conjointly, or by so many distinct letters. Thus, so many separate equations will be obtained, the resolution of which, by some of the foregoing rules, will lead to the determination of the quantities required.


1. Given the sum of two magnitudes, and the difference of their squares, to find those magnitudes separately.

Let the given sum be denoted by s, the difference of the squares by D; and let the two magnitudes be represented by x and y respectively.

Then, the first condition of the problem expressed algebraically is x + y = s

And the second is x2

ya = D

Equa. 2 divided by equa. 1, gives x — y =

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2 S

2 s

To apply this general solution to a particular example, suppose the sum to be 6, and the difference of the squares 12.

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36 +12

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2 s

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Ex. 2. Given the product of two numbers, and their quotient, to find the numbers.

Let the given product be represented by p, the quotient by q; and the required numbers by x and y, as before.

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Extracting the square root, y =

Then, by substitution, aqy = q

✔p q.


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Suppose the product were 50 and the quotient 2.


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✓100 = 10.

Again, suppose the product 36, and the quotient 21.

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Ex. 3. Given the sum (s) of two numbers, and the sum of their squares s, to find those numbers.

Ans. x = 1 s + √2s-s2, and y = s-√2 s — s2.

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Ex. 4. The sum and product of two numbers are equal, and if to either sum or product the sum of the squares be added, the result will be 12. What are the numbers?

Ans. each

= 2.

Ex. 5. The square of the greater of two numbers multiplied into the less, produces 75; and the square of the less multiplied into the greater produces 45. What are the


Ex. 6. A man has six sons whose successive ages differ by four years, and the eldest is thrice as old as the youngest. Required their several ages?-Ans. 10, 14, 18, 22, 26, and 30 years.

SECTION IX.-Quadratic Equations.

When, after due reduction, equations assume the general form A x2 + в x + c = 0; then dividing by A, the coefficient

of the first term, there results x2 +

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x +

= 0, or, making

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9 we have x2 + px + q = 0



an equation which may represent all those of the second degree, p and q being known numbers positive or negative.

Let a be a number or quantity which when substituted for x renders x2+p x+q=0; then a2+pa+q=0, or q――a3-pa. Consequently+px+q, is the same thing as x2-a2+p px-pa, or as (x+a) (x − a) + p (x—a), or, lastly, as (x—α) (x + a +p).

The inquiry, then, is reduced to this, viz. to find all the values of which shall render the product of the above two factors equal to nothing. This will evidently be the case when either of the factors is = : 0; but in no other case. Hence, we have x―a,=0, and x+a+p=0, or x=a, and x ——— a — p.*

And hence we may conclude

1. That every equation of the second degree whose conditions are satisfied by one value a of x, admits also of another value-a-p. These values are called the roots of the quad

ratic equation.

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2. The sum of the two roots a and-a-p is-p; their product is a2 - a which as appears P, above is q. So that the coefficient, p, of the second term is the sum of the roots with a contrary sign; the known term, q, is their product.

3. It is easy to constitute a quadratic equation whose roots shall be any given quantities b and d. It is evidently a2 (b+d) x+b d=0.

4. The determination of the roots of the proposed equation (1) is equivalent to the finding two numbers whose sum is-p, and product q.

5. If the roots b and d are equal, then the factors b and xd are equal; and x2+p x+q is the square of one of them.

To solve a quadratic equation of the form x+px+q=0, let it be considered that the square of x + p is a trinomial, x2+ px+p2, of which the first two terms agree with the first two terms of the given equation, or with the first member of that equation when q is transposed.

That is, with x2+px=—q.
Let then

be added, we have, x+px+ } p2=\p2 — q of which the first member is a complete square.

If it be affirmed that the given equation admits of another value of x, besides the above, b for instance, it may be proved as before that x- b must be of the number of the factors of x2+p x+q, or of (x− a) (x+a+p). But x-a and x+a+p being prime to each other, or having no common factor, their product cannot have any other factor then they. Consequently 6 must either be equal to a or to-a-p; and the number of roots is restricted to two.

Its root is x+ p=± √(} p2 — q)

and consequently p± √(‡p2—q)
x = — }
otherwise, from number 2 above, we have x + x' =
and x x' = q



Taking 4 times the second of these equations from the square of the first, there remains x2 · 2 x x'+x'2=p2 x

Whence, by taking the root, — x'= √(p2 — 4 q)
Half this added to half equa. 1, gives x=


} √(p2 - 4 q) = − } p + √(4 p* — q)
And the same taken from half equa. 1, gives x':
√(p3 — 4 q)




✔(4 po — q) which two values of

a evidently agree with the preceding.

It would be easy to analyze the several cases which may arise, according to the different signs and different values, of p and q. But these need not here be traced. It is evident that whether there be given

1. x2+p x=q

2. x2-p x=q
3. x2+p x=— - Չ

4. x2—p x= — 9

The general method of solution is by completing the square, that is, adding the square of p, to both members of the equation, and then extracting the root.

It may farther be observed that all equations may be solved as quadratics, by completing the square, in which there are two terms involving the unknown quantity or any function of it, and the index of one double that of the other. Thus,

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x.6 ± p x3 = q, x2 ±p x2 = q, x2± p x1 = q, (x2 + p x + q)3 + (x2+p x+q)=r, (x2n — x1)3 ± (x2-x")=q, &c., are of the same form as quadratics, and admit of a like determination of the unknown quantity. Many equations, also, in which more than one unknown quantity are involved, may be reduced to lower dimensions, by completing the square and reducing; such, for example, as (23+y3)3±p(z3+y3)

so on.

хо px = 9, ± = 9, and y


Note. In some cases a quadratic equation may be conveniently solved without dividing by the coefficients of the square, and thus without introducing fractions. To solve the general equation a x2 + bx = c, for example, multiply the whole by 4 a, whence 4 a2x2 + 4 a b x = 4 a c, adding b2 to complete the square, 4 a2x2 + 4 a b x+b2=4 a c+b2 taking the square root, 2 a x + b = ± √(4 a c+b3) ;

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