Completing the square, x^2 — 7 x + ({7}{2})^2 = 8 + {49}{4}
that is, (x - {7}{2})2 = {81}{4} ;
x - {7}{2} = \pm {9}{2} = {7}{2} \pm {9}{2} = 8, or - 1.
Note. We do not work out ({7}{2})^2- on the left-hand side.
Ex. 2. Solve 32 -3x^2 = 10x.
Transposing, 3 x^2 + 10 x = 32. Divide throughout by 3, so as to make the coefficient of x^2 unity. Thus x^2 + {10 }{3}x = {32}{3}. completing the square, x^2 + {10 }{3}x +( {5}{3})^2= {32}{3} +{25}{9} that is (x + {5}{3})^2 = {121}{9}; x + {5}{3}= \pm {11}{3} x = - {5}{3} \pm {11}{3} =2, or -5{1}{3}.
Ex.3. Solve 7(x + 2a)^2 + 3a^2 = 5a(7x+ 23a).
Simplifying, 7 x^2 + 28 ax + 28 a^3 + 3 a^2 = 35ax + 115 a^2 , that is, 7 x^2 — 7 ax = 84 a^2 ; Whence x^2 — ax = 12 a^2;
completing the square, x^2 — ax + ({a}{2})^2 = 12 a^2 + {a^2}{4}; that is, (x - {a}{2})^2 = {49a^2}{4} x - {a}{2} = \pm {7a}{2} x = {a}{2} \pm {7a}{2} x = 4a or - 3a
286. We see then that the following are the steps required for solving an affected quadratic equation.
(1) If necessary, simplify the equation so that the terms in x^2 and x are on one side of the equation, and the term without x on the other.
(2) Malce the coefficient of x^2 unity and positive by dividing throughout by the coefficient of x^2.
(3) Add to each side of the equation the square of half the coefficient of x.
(4) Take the square root of each side.
(5) Solve the resulting simple equations.
287. The quadratic equations considered hitherto have had two roots. Sometimes, however, there is only one solu-
