Title Target-1_Level-1_Chapter-1_NoRestriction.pdf ✅

1 Quadratic Equations CHAPTER 1 LEVEL-I SOLUTIONS [Properties Related to Nature of Roots] 1. Answer (1) For an identity (k2 – 3k + 2) = 0 ⇒ (k – 1) (k – 2) = 0 ⇒ k = 1, k = 2 k2 – 5k + 4 = 0 ⇒ (k – 1)(k – 4) = 0 k = 1, 4 k2 – 6k + 5 = 0 ⇒ (k – 5)(k – 1) = 0 k = 1, 5 Common value of k = 1. 2. Answer (4) ( )( ) ( )( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) a ba c a x b x c b cb a b x c x a c ac b c x a x b x is satisfied by x = a, x = b, x = c. A quadratic equation is satisfied by more than two values of x. So it is an identity. Hence it is satisfied by all values of x. 3. Answer (1) x 3 (x 7x 10 0 ) 2 ⇒ x = 3 and x = 2, 5 But x ≥ 3 ∴ Only two roots. 4. Answer (2) x3 – 2x2 – x + 2 = 0 As x = 1 is the root of the equation
2 Quadratic Equations Solutions of Assignment (Level-I) Hence we may write x3 – 2x2 – x + 2 = x2 (x – 1) – x(x – 1) – 2(x – 1) = (x – 1) (x2 – x – 2) = (x – 1) (x – 2) (x + 1) Roots = 1, –1, 2. 5. Answer (2) If 1, 2, 3 are roots of equation then x3 + ax2 + bx + c = 0 ⇒ 1 + 2 + 3 = –a ⇒ a = –6 1⋅2 + 2⋅3 + 1⋅3 = b ⇒ b = 11 1⋅2⋅3 = –c ⇒ c = –6 6. Answer (2) , , q p l n l n Now, q p p q l n l n l n q p p q l n q p p q l n & 0 7. Answer (1) (b c a x) (c a b x) (a b c) 0 2 Put x = 1, bcacabab c a b c 0 ∴ 1 is the root of the equation. ∴ Roots are rational. 8. Answer (2) Let p 4p m 2 2 as x p p p 2 4 2 ! is an integer ⇒ p 4p m 0 2 2 ⇒ p 2 4 m2 ! ⇒ 4 m2 + is an integer Let 4 m2 2 where ! I ⇒ m 4 2 2 ⇒ !2,m 0 It follows that p2 – 4p = 0 ⇒ p(p – 4) = 0 for p = 0, 4

4 Quadratic Equations Solutions of Assignment (Level-I) 14. Answer (3) Let y mx m9 5m 1 2 We need y > 0 ⇒ Upward parabola above x-axis. mx mx 9 5m 1 0, x R. 2 6 ! ⇒ D < 0, a > 0 i.e., 81m m4( )(5m 1 0 ) < and m > 0 2 ⇒ m m (61 4 0 ) and m 0 0 m 61 4 − < > & < < Also for m = 0, 0x 9( ) 0 x 0 1 1 0> , x R 2 6 ! ∴ m 0, 61 4 ! : k [Location of Roots] 15. Answer (2) x x 6 0 x x 6 0 2 2 & ⇒ ^ x 3h^ x 2h 0 & x 3, x 2 ⇒ x = ± 2 Two roots are real, with sum 0. [Polynomial Equations and Transformation of Roots] 16. Answer (3) p + iq is one root ⇒ p – iq is other root. Let α be third root. Now sum = α + p + iq + p – iq = 0 ⇒ 2p 2p is root of x ax b 0 3 ∴ 2p is root of ( x) ax b 0 3 ⇒ x ax b 0 3 17. Answer (3) ax bx c 0 2 , Given equation is an2 – bx (x – 1) + c(x – 1)2 = 0 a x x b x x 1 1 1 0 2 ` j ` j Now, Replacing x by x x 1 ( ) ( ) ( ) x ax x bx c ax bx x c x 1 1 0 1 1 0 2 2 2 2 & x x x 1 1 & is the root of the above equation.