to be incomplete: secondly, it may happen that the coefficient of the first power of the unknown quantity is not equal to 0; in this case the equation is said to be complete. Incomplete equations, when reduced, contain but two terms one containing the square of the unknown quan. tity; the other, a known term. Complete equations, when reduced, contain three kinds. of terms viz.: a term containing the square of the unknown quantity, a term containing the first power of the unknown quantity, and a known term. First Case. Incomplete Equations. 143. In this case, the reduced equation takes the form, x2-= q. Extracting the square root of both members, we have, x = ±√9. Hence, the rule for solving incomplete equations: RULE. Reduce the equation to the form, x2: =q, and extract the square root of both members. There will be two roots numerically equal, but having contrary signs. Denoting the first root by ', and the second by x", we have, 5. 2. 3x2 4 28+ x2. 6. 7 4. x2 + ab = 5x2. 8. 3x2 + 5 EXAMPLES. x 2 (x + α)2 =2ax + b. x2 + 29 x + 7 2+2 16. = 117 Ans. Ans. x' = + √b — a2, x'' = x√α + x2 = b + x2. Ans. x + Ans. x' = + 3, x" = Ans. x = + 4, x2' = = Ans. x' = +√ab, x" = − + √ab, Va x2-73 Ans. x = + 9, 5x2. = +5, x'' = √3, x' = a - 3. 48 vb - a2. = - 4. - 5. b 9. 26 - v Second Case. Complete Equations. 144. The reduced form of the complete equation is, x2 + 2px = 9. Adding p to both members (Axiom 1), we have, q + p2. x2 + 2px + p2 Extracting the square root of both members (Axiom 5) we have, = x + p = ± √ï + p2. Transposing p to the second member, we have, x = − p ± √ q + p2. Hence, there are two roots, one corresponding to the plus sign of the radical, and the other to the minus sign. Denoting these roots by x' and ', we have, x = − p + √ q + p2, and 2" = − p − √q + p2. Hence, we have the following rule for solving complete equations of the second degree: RULE. I. Reduce the equation to the form, x2 + 2px = 9, by the rule. II. The first root is equal to half the coefficient of the second term, taken with a contrary sign, plus the square root of the second member increased by the square of half the coefficient of the second term. III. The second root is equal to half the coefficient of the second term, taken with a contrary sign, minus the square root of the second member increased by the square of half the coefficient of the second term. 1. Let it be required to solve the equation, 3x2 14x+15= 0. 3 X 5 + 7 + 2 49 9 = 14 3 4x x= + and x"= 3, and x"= - 5. 5 3 These roots may be verified. Substituting 3 for x, in the given equation, we have, 3 × (3)2 - 14 × 3 + 15 = 0; or, 0 = 0. 5 Substituting for x, in the same equation, we have, 3 9 718 713 which shows that both 3, and are roots. Writing out the roots, by the rule, Vi 81 5 14 X +15=0; or, 0 = 0: x = 7. = 14. and x" = 213 2. Again, let it be required to solve the equation, 14 x x + 1 Clearing of fractions, and reducing to the required form, x2 = 9 5+ 49 Or, reducing, 9 25 x' = + = 4, 8 8 which roots may be verified as before. 3. Given 3x + 4 x2 48x= Writing out the two roots, x = 24+ √-432 + 576, Or, reducing, and x" = 30 2x x 6 6. ax2 - bx = c. - Ans. x = 36, =3 Solve the following equations: 4. 5.2 - 6x60 3. 5. (12) (x+2)= 0. 7x 10 Clearing of fractions, and reducing to the required form we have, 432. 9 b + √b2 + 4αc and x" 12. In writing out the roots by the rule, it frequently happens that the quantity under the radical sign is made up of two fractions, or of an entire part and a fraction, as in Examples 1 and 2. In such cases, the two parts must be reduced to the least common denominator which is a perfect square, and then added together, and reduced to the simplest form. Ans. x' = 23 8 and x 24-√-432+576 21 b x = " = 9 5 Ans. x' = 12, x' = to find a √b2 + 4ac |