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EX 3.1 2.Determine the nature of the roots of kx2 – x + k = 1 –2kx, where k 0 , and find the roots in terms of k, where necessary Solutions: kx2 +2kx – x + k – 1 = 0 kx2 + x( 2k –1 ) - 1 = 0 a=k b = 2k – 1 c=-1 D = b2 – 4ac = ( 2k – 1 ) 2 – 4 (k) (k- 1) = 4k2 – 4k +1 - 4 k2+4k = 1 which is > 0 Solutions are x b D 2a 3.Find the value(s) or range of values of p for which the equation a) px2 – 6x +p = 0 has equal roots, a)Since the equation has real and equal roots , the discriminant , D = 0 a=p b= –6 c=p D = b2 – 4ac =( - 6 ) 2 – 4 (p) (p) = 36 – 4p2 D= 0 i.e 36 – 4p2 36 p2 p =0 = 4p2 =9 =3 3b) 2x2 – 4x +3 = p has real roots Since the equation has real roots , the discriminant D 0 Rewrite the given equation as follows 2x2 – 4x +3-p =0 a=2 b= –4 c = 3- p D = b2 – 4ac 0 i.e (-4)2 – 4 (2 ) (3-p) 0 16 – 24 + 8p 0 -8 +8p 0 8p 8 p 1 11.If the roots of the equation px2 +qx +r = 0 are real , show that the roots of the equation r2x2 + (2pr – q 2 )x +p2 = 0 are also real. Since the equation px2 +qx +r = 0 has real roots , the discriminant D 0 a=p b= q c=r D = b2 – 4ac 0 i.e (q)2 – 4 (p ) (r) 0 q2 – 4p r 0 -----------------------(i) Consider the equation r2x2 + (2pr – q 2 )x +p2 = 0 a = r2 b = 2pr – q 2 c = p2 = b2 – 4ac = (2pr – q 2)2 – 4 (r2)(p2) = 4 r2p2 –4pq2r +q4– 4 r2p2 = –4pq2r +q4 = q2 (q2 – 4p r) 0 ( because q2 0 for any value of q and q2 – 4p r 0 ) So, the equation px2 +qx +r = 0 also has real roots. D 13 From the simultaneous equations 2x – 3y = 4 , 2x2 – 9y2 =k. Derive an equation relating x and k. Hence find the range of values of k if the simultaneous equations have real solution(s). Solution: 2x – 3y = 4 -------------(1) 2x2 – 9y2 =k.------------(2) From Eqn. (1) 3y = 2x –4 Substitute in eqn (2), we get 2x2 – ( 2x-4)2 2x2 – (4x2 –16x +16 ) – 2x2 +16x -16 ) 2x2 –16x +16 +k =k =k =k =0 Since the equation 2x2 –16x +16 + k = 0 has real roots , the discriminant D 0 a=2 b = -16 c = 16+k D = b2 – 4ac 0 i.e ( -16)2 – 4 ( 2 ) ( 16 +k ) 256 –128 -8k 128 -8k -8k k 0 0 0 -128 16 13.Given that x(x-2) = t – 2 , find x in terms of t. Hence , find the range of values of t for x to be real. The Given equation can be written as x2 – 2x –t +2 = 0 Since a = 1, b = - 2 , c = 2 – t D = b2 – 4ac =( -2 )2 – 4 ( 1) ( 2- t) =4 – 8 +4t =-4+4t x = 1 sqr (t-1) x 1 t 1 For the equation x2 – 2x –t +2 = 0 has real roots , the discriminant D 0 i.e -4+4t 4t t 0 4 1