2. EX 3.1 2.Determine the nature of the roots of kx 2 – x + k = 1 –2kx, where k  0, and find the roots in terms of k, where necessary Solutions: kx 2 +2kx – x + k – 1 = 0 kx 2 + x( 2k –1 ) - 1 = 0 a = k b = 2k – 1 c = - 1 D = b 2 – 4ac = ( 2k – 1 ) 2 – 4 (k) (k- 1) = 4k 2 – 4k +1 - 4 k 2 +4k = 1 which is > 0 Solutions are

3. 3.Find the value(s) or range of values of p for which the equation a) px 2 – 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 = b 2 – 4ac =( - 6 ) 2 – 4 (p) (p) = 36 – 4p 2 D= 0 i.e 36 – 4p 2 = 0 36 = 4p 2 p 2 = 9 p =  3

4. 3b) 2x 2 – 4x +3 = p has real roots Since the equation has real roots, the discriminant D  0 Rewrite the given equation as follows 2x 2 – 4x +3-p =0 a = 2 b = – 4 c = 3- p D = b2 b2 – 4ac 00 i.e (-4) 2 – 4 (2 ) (3-p)  0 16 – 24 + 8p 00 -8 +8p 00 8p  8 p 11

5. 11.If the roots of the equation px 2 +qx +r = 0 are real, show that the roots of the equation r 2 x 2 + (2pr – q 2 )x +p 2 = 0 are also real. Since the equation px 2 +qx +r = 0 has real roots, the discriminant D  0 a = p b = q c = r D = b 2 – 4ac  0 i.e (q) 2 – 4 (p ) (r)  0 q 2 – 4p r  0 -----------------------(i)

6. Consider the equation r 2 x 2 + (2pr – q 2 )x +p 2 = 0 a = r 2 b = 2pr – q 2 c = p 2 D = b 2 – 4ac = (2pr – q 2 ) 2 – 4 (r 2 )(p 2 ) = 4 r 2 p 2 –4pq 2 r +q 4 – 4 r 2 p 2 = –4pq 2 r +q 4 = q 2 (q 2 – 4p r)  0 ( because q 2  0 for any value of q and q 2 – 4p r  0 ) So, the equation px 2 +qx +r = 0 also has real roots.

7. 13 From the simultaneous equations 2x – 3y = 4, 2x 2 – 9y 2 =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) 2x 2 – 9y 2 =k.------------(2) From Eqn. (1) 3y = 2x –4 Substitute in eqn (2), we get 2x 2 – ( 2x-4) 2 = k 2x 2 – (4x 2 –16x +16 ) = k – 2x 2 +16x -16 )= k 2x 2 –16x +16 +k= 0 Since the equation 2x 2 –16x +16 + k= 0 has real roots, the discriminant D  0

8. a = 2 b = -16 c = 16+k D = b 2 – 4ac  0 i.e ( -16) 2 – 4 ( 2 ) ( 16 +k )  0 256 –128 -8k  0 128 -8k  0 -8k  -128 k  16

9. 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 x 2 – 2x –t +2 = 0 Since a = 1, b = - 2, c = 2 – t D = b 2 – 4ac =( -2 ) 2 – 4 ( 1) ( 2- t) =4 – 8 +4t =-4+4t x = 1  sqr (t-1)

10. For the equation x 2 – 2x –t +2 = 0 has real roots, the discriminant D  0 i.e-4+4t  0 4t  4 t  1