Extracting the square root, y = Then, by substitution, x = q y = q ✔p q. Suppose the product were 50 and the quotient 2. Again, suppose the product 36, and the quotient 21. Ex. 3. Given the sum (s) of two numbers, and the sum of their squares s, to find those numbers. 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 numbers? 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 + B x + c = 0; then dividing by A, the coefficient of the first term, there results x2 + x + = B C p = 9 = Α' A B A A O, or, making we have x2 + px + q = 0......(1) 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 —p a. Consequently x2 + px+q, is the same thing as 22. -a2 + p px-pa, or as (x+a) (x-a) + p (x-a), or, lastly, as (x-a) (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 a,=0, and x+a+p=0, or x=a, and x =— a — p.* And hence we may conclude have x 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 quadratic equation. -- 2. The sum of the two roots a and-a-p is—p; their product is a9 a p, which as appears 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 x2 (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. x b and 5. If the roots b and d are equal, then the factors x d are equal; and x2+p x+q is the square of one of them. To solve a quadratic equation of the form x+p x+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= Let then p2 be added, we have, x+px+ p2={p3 — q of which the first member is a complete square. x, b must be of the num If it be affirmed that the given equation admits of another value of besides the above, b for instance, it may be proved as before that ber 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) = P and xx' = 9 Taking 4 times the second of these equations from the square of the first, there remains 2 · 2 x x'+x'2=p2 — 4 q Whence, by taking the root, x-x'= √(p3 — 4 q) Half this added to half equa. 1, gives x= } p+ √(4 p3 — q) } √(p2 — 4 q) == - 1 p + And the same taken from half equa. 1, gives x': } √(p3 — 4 q) - P - ≤ p − √ ( ‡ p2 -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 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. 22 2 Thus, x3 ± p x3 = q, x2n ±p x" = q, xa3± pxa = q, (x2 + p x + q)2 ±(x2+p x+q)=r, (x2n — x1)2 + (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 (x3+y3)3±p(z3+y3) so on. x2 px = 9, and 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 a3 x2 + 4 a b x = 4 a c, adding bo to complete the square, 4 a2x2 + 4 a b x+b2=4 a c+bo taking the square root, 2 a x + b = ± √(4 a c+bo); transposing the 10, x2. 8x= 1910-9 completing the squ., x2-8x+16=9+16=25 extracting the root, x-4-5 1. to find the values of x. dividing by 23, 22 — 4 x = x2 - 14, 6 3 1 22 x2 구름을 9 complet. squ. 2-1 1 x + (17)2 = 721-11-11 3. Given x2+2 x +4 √x2+2x+1=44, to find x. adding 1, we have (x2 + 2 x+1) + 4 √(x2 + 2 x + 1) = 45 = 49 9 complet. squ. (x2 + 2 x + 1) + 4 √(x2 + 2 x + 1)+4 transposing the 2(x+2x + 1) = ± 7 − 2 = 5 01 Substituting the former value of x in the 2d equa. it becomes 2 y-y-2, or y=2; whence x=4. Again, substituting the 2d value of x, in equa. 2, it becomes whence = y - 2, and x = + 12. 6 y — y or 7 y = 2; 6. Given x2 y3 54 x y, and x and y. equa. 1, by transposition, becomes a y2-4 x y=5 extracting the root, x y―2=±3 Whence x y=5 or 1. Substituting the first of these values for x y in equa. 2, it becomes y3: whence y=1 and x=5. Substituting the 2d value in the same equation, it becomes ✔} = — } 25, and x = - 1 = - whence y -25= 5. = s. A man travelled 105 miles at a uniform rate, and then found that if he had not travelled so fast by two miles an hour, he would have been six hours longer in performing the same journey. How many miles did he travel per hour? Ans. 7 miles per hour. 9. Find two such numbers that the sum, product, and difference of their squares may be equal. Ans. §+ √5, and 2+ √5. 10. A waterman who can row eleven miles an hour with the tide, and two miles an hour against it, rows five miles up a river and back again in three hours: now, supposing the tide to run uniformly the same way during these three hours, it is required to find its velocity? Ans. 412 miles per hour. |