x + y (a - x)! 1 (1+ 3/5-5 3y3 (ax) 2-3y In the following examples simplify each fraction separately before reducing them to a common denominator. 12. Simplify the following expressions : 13. Extract the square roots of the following binomial surds :2+√√3; 11−6√√2; 17-12√2; 63+4√143; + √√2; 9 4 8-2√7; 43 2 2 (10-3/11); 28+ 8/10; 99 23 8 + 37√3 – 60; 11√7+2√210; √ 2(a2+b2 + ab + (a + b) √√ a2 + b2); 2(a+1)(a - √√a2-1); 2 (a2 + b2 + c2 + (a + b + c) √ a2 + b2 + c2). 14. Find the fourth root of 17-12√2; 92 + 32 √7; § (73+28√6); 33 – 20√2; 14 √3+ 24. 15. Find the cube root of 584+ 195√√/3; of 45 – 29 √2; the eighth root of 577-408/2; the ninth root of 109. An equation is called Quadratic when the highest power of x contained in it is the second or square. Thus and 3x2 = 75 (1.) (2.) are both quadratic equations. 110. An equation in which only the second power of x appears is called a "pure" Quadratic. x = +5 x = -5 for +5 and 5 are each a square root of 25. It is usual to write the result of the square root thus: 111. To solve a pure Quadratic Equation, observe the following RULE: (a.) Clear the equation of fractions if necessary. (b.) Collect all the terms involving x2 on the first side of the equation, and the known quantities on the second side, and simplify. (c.) Divide both sides by the co-efficient of x2. (d.) Extract the square root of both sides, and prefix the double sign. Examples (LII.) 7x2-2 = 26 ; 1−x2 = 4x2−24; 2 (1+x2) = 7 (1−x2) +4 8+3x2−(11+2x2) = 0; 2 (x2+1) +3 (x2+2) +4 (x2+3) −2 = 0 (x+a) (x−b) (x−c) + (x − a) (x+b) (x−c) + (x−a) (x−b) (x+c) = (x+a)(x+b)(x−c)+(x+a) (x−b) (x+c) + (x− a)(x+b)(x+c) 112. An equation in which both the first and second powers of x occur is called an adfected quadratic. Before giving examples of the solution of adfected quadratic equations, it will be necessary to discuss the process of "completing the square." is a perfect square, and it is seen that the last term (a2) is the square of half the coefficient of x, and that the coefficient of (x2) is +1. Now, suppose we take the expression x2+6x; if it be required to make this a complete square, we must add the square of half the coefficient of x, viz. (3)2; it then becomes x2+6x+9, which is the exact square of x+3. Similarly to make x2+10α2x a complete square, we must add (5a2)2 or 25a4. Then x2+10a2x+25a1 = (x+5a2)2. Again, to make x2- px, a complete square, p2 must be added, and the square is Hence then we may complete the square when the expression contains 2 (the coefficient being +1) and x with any coefficient by the following RULE. Add to the two given terms the square of half the coefficient of x. Examples LIII. Complete the squares in the following expressions: 9 (1.) x2+4x; x2-6x; x2+x; x2+3x; x2+ x; x2+ 4 x2+ax; x2−3bx; x2-8abx; x2-3px; x2−2α2x; x2+9p3x. (2.) Complete the square in the expression Note. Whenever the index of the letter in one of the terms is double of its index in the other term of a binomial, the square may be completed by the above Rule, as in the instance just given. (3.) Make the following expressions complete squares : x1+x2; x¤—ɑx3; x2—a2x2; xa — a2x2; x1o–13×3; x6+10p2q2x3 (4.) Write down the "square root" of each of the completed squares in the above examples. Let it be required to solve the equation 3x2-12x+4=-5 Working. By transposition 3x2-12x = -9 Dividing both sides by 3 x2-4x= -3 Completing the square by adding (-2)2 or 4 to each side, Hence 3 and 1 are the two roots of x in the above equation. Again to solve the equation ax2+bx-c = 0 Proceeding by the same steps as above, we have These values may be put in the more convenient form x = b-4ac+b2 2a b+4ac+b2 x = |