a) Q is the common root of the two equations Therefore, 3Q^2+aQ+b=0 ......(1) and 3Q^2+bQ+a=0 ......(2) (1) - (2): (a-b)Q + (b-a) = 0 Q= 1 b) put Q=1 into (1), we have a+b+3 = 0 b= -(a+3) a and b are the roots of x^2+hx+k=0 Therefore, h = -(a+b) and k = ab h = -(a+b) = -[a - (a+3)] = 3 k = ab = -a(a+3) > 0 => -3 < a < 0 Since k is a positive integer, a must be equal to -1 or -2 which both gives k=2 Remarks Actually, (a,b) = (-1,-2) or (-2,-1) gives the same two equations in part (a) [ 本帖最後由 kazuhiko 於 4/10/2006 00:17 編輯 ] |