.. X = 56 find x. 99' seen (Stage I., Alg.; Art. 35) that the unknown quantity in a pure quadratic has always two values, equal in magnitude, but of opposite sign. Ex. 1. Given 3 x + 12 687, find x. We have 3 x 687 12 = 675, or ca 225. † 15. 2 + 7 2x - 7 Ex. 2. Given 2." - 730 22€ + 7 x 6x2 Bringing the fractions on the first side to a common denominator, we have (2 x + 7)2 – (2 x 7) 56 56 99' 1 49 6 sc* 99, clearing of fractions ; then, 6 99 4 22 – 49, from which 25 56 aC or, or, 4 cm Adfected Quadratics. 3. Solution by completing the square. Suppose we have given the equation 202 + 2 ax = 3 a’, to find 3. It will be remembered that (x2 + 2 ax + a?) is a perfect square, viz., the square of (x + a). It is evident, then, that the first side of our equation will become a perfect square by the addition of aas a third term. Adding, then, a’ to each side of the given equation, we haveacea + 2 ax + a? 4 a, or (x + a)? 4a. Taking now the square root of each side, we have I 2 a. ..a + 2α + a = a. a or 3 a. We may remark that the quantity a, added to the expression x2 + 2 ax in order to make it a perfect square, is the square of half the coefficient of x. The operation itself is called completing the square. An adfected quadratic may therefore be solved as follows: 1. Reduce and arrange it until all the terms involving x are on the first side, and the coefficient of xo is unity. 2. COMPLETE THE SQUARE by adding to each side the SQUARE OF HALF the coefficient of x. 3. Take the square root of each side, put a double sign to the second side, and transpose the term of the first side not involving x. Ex. 1. Solve the equation 3 x2 + 18 + 4 52. We have 3 x2 + 18 x 52 4 48; or, dividing each side by 3, ga + 6x 16. Here, the coefficient of x is 6, the half of which is 3. Adding then the square of 3 to each side, to complete the square, we have ocia + 6 x + 32 16 + 9 = 25. Taking the square root of each side, we have x + 3 = † 5. 42c + 9 12 x + 17 Ex. 2. Given 2x + 7 6x + 16 find x. - (2 - 6x + 16) ; 1 1 2 2x + 7 15 6x + 16) 1 1 15 or, simplifying; X + 4 6x + 16 7 7 15 (2 x + 7) - 5 (6 x + 16), 25 ôr, simplifying, ca + 63 + 8 12x + 74 x + 112 x + 2 2 X + 24; 4 or or, clearing of fractions 24 2a + 148 x + 224 = 252c + 150 x + 200; or, transposing and reducing, x2 24; or, changing the sign of each side, then ca + 2x or, completing the square, ac + 2x + 24 + 1 = 25. Taking the square root of each side, we have— X + 1 = +5 † 5 1 6. Ex. 3. Solve the equation ac + 6 x + 25 = 0. We have xa + 6 - 25; or, completing the square cca + 6 + 32 = 25 + 9 16; or, extracting the square root of each side ť - 16 ii X = 3 + N = 16. As the quantity ✓ - 16 has no exact or approximate value, the given equation has no real roots. The roots are therefore said to be imaginary. 4. Solution by breaking into factors. We have seen (Stage I., Alg., Art. 30) that it is often easy to find by inspection the factors of quadratic expressions. We e may make use of this knowledge to solve quadratic equations. Ex. 1. Solve the equation x2 + 5x = 66. x2 + 5x 66 0: And, resolving the first side into its elementary factors, we get (x – 6) (2 + 11) = 0. Now, if either of these factors is put equal to 0, the equation is satisfied. Hence, we have, s 6 0, and x + 11 0; 6, and x = 11. : 6 and 11 are the roots required. or, X = Ex. 2. Given - (a + b)2 + ab = 0, find z. Now this equation is satisfied by making either of the factors = 0. Hence, - Q = 0, or s = a; and, is b = 0, or ac = b;) :. a, b are the roots required. 1 1 10. = ax. Vatica 1 ax 4 Na2x2 Va* + acid + Va? zca 11. Var + Va? - 2 a + Vac 1 12. 13. 22 5x + 6 = 0. Na . 1 14. 72. 15. 3 x2 X = 2. 16. 42 28. 0. 18. **+ 63 2 + 8 = 0. 19. ax + b + c = 0. 14 42 20. 3 = 42 - 7. 1. 3 26 22. 2 x + 1 2x + 1 9 w = 35 = 21. a = 11. See Remark, page 304. 24. (a 23. (a + bx) (cx + d) = (a + b) (c + dx). (a - b) (22 – 62) (a + b) (20 – 6). 25. (a + b) (x - a) (x 6) abx. 6 a? 62 26. b ab a + 2 + cd 30. = SC + 33. + 27. axa (c + d) x = bx? CV 7 28. axa (a 6)2 x + a3 63202 + (a? + ab + 6?)?x + 69. 23 2 29. 3. 7 2x + 5 2 2 + 9 31. 53. 10 32. 0. 2 + 3 X + 8 7 5x + 3 5 2 7x + 4 9 6x + 5 4 1 34. 7 X + 1 2 2 3 7 X + 8 4 x + 13 32 2 2 + 9 2 x + 24 36. 2 22 1 2 x + 14 x 9* X + 1 39. * 15x + 6 + Noc + 10 4. 40. 13x 4 12 x 4 41. Va? + b2 + x + Na+ + 35. C |