Title 02 Unit # 2 (38-86) ok.pdf ✅

YOUTH ACADEMY YOUTUBE CHANNEL SIR USAMA Youth Academy Official Website: www.youthacademy.pk Discriminant (BWP 2018) (U.B + K.B) “For a standard quadratic equation 2 ax bx c + + = 0 , the value of the expression 2 b ac − 4 is called discriminant.” It is used to find the nature of roots without solving the equation. Nature or Characteristics of the Roots (U.B + K.B) Nature of a quadratic equation 2 ax bx c + + = 0 , when a b c Q , ,  and a  0 as: (i) If 2 b ac − = 4 0 , then the roots are rational (real) and equal. (ii) If 2 b ac −  4 0, then the roots are complex conjugate or imaginary. (iii) If 2 b ac −  4 0, and is a perfect square, then the roots are rational (real) and unequal. (iv) If 2 b ac −  4 0, and is not a perfect square, the roots are irrational (real) and unequal. Note (K.B) If given polynomial expression is a perfect square then discriminant is 0. Example 2: (Page # 19) Using discriminant, find the nature of the roots of the following equation and verify the result by solving the equation. 2 x x − + = 5 5 0 (LHR 2015, GRW 2016, 17, SWL 2017, RWP 2015, D.G.K 2017) Solution: 2 x x − + = 5 5 0 Here a b c = = − = 1, 5, 5 Discriminant = 2 b ac − 4 2 = − − ( 5) 4(1)(5) = − = 25 20 5 As discriminant > 0 but not perfect square, Roots are irrational (real) and unequal. Verification: Solving the equation by using quadratic formula 2 4 2 b b ac x a −  − = 2 ( 5) ( 5) 4(1)(5) 2(1) x − −  − − = 5 5 2 x  = Evidently, Roots are irrational (real) and unequal. Example 2: (Page # 21) Find k, if the roots of the equation 2 ( 3) 2( 1) ( 1) 0 k x k x k + − + − + = are equal, if k −3 (A.B) Solution: 2 ( 3) 2( 1) ( 1) 0 k x k x k + − + − + = Here a k b k c k = + = − + = − + 3, 2( 1), ( 1) As roots are equal, discriminant is zero. 2  = − = Disc. 4 0 b ac 2 [ 2( 1)] 4( 3)[ ( 1)] 0 − + − + − + = k k k 2 4( 1) 4( 3)( 1) 0 k k k + + + + = 4( 1) ( 1) ( 3) 0 k k k + + + + =   4( 1) 2 4 0 k k + + = ( ) Either k + =1 0 or 2 4 0 k + = 4 0  k =−1 or 2 4 k =− k =−2 Thus, roots will be equal if k = − − 1, 2 Exercise 2.1
YOUTH ACADEMY YOUTUBE CHANNEL SIR USAMA Youth Academy Official Website: www.youthacademy.pk Q.1 Find the discriminant of the following given quadratic equations: Solution: (i) 2 2 3 1 0 x x + − = (A.B) (GRW 2017, FSD 2016, MTN 2014, D.G.K 2016) By comparing given equation with 2 ax bx c + + = 0 , we get 2, b 3, c 1 a = = = − Disc = 2 b ac − 4 = 2 (3) 4(2)( 1) − − = 9 + 8 = 17 (ii) 2 6 8 3 0 x x − + = (A.B) (LHR 2016, SWL 2016, D.G.K 2015, 17) By comparing given equation with 2 ax bx c + + = 0 , we get 6, 8, 3 a b c = = − = Disc = 2 b ac − 4 = 2 ( 8) 4(6)(3) − − = 64 – 72 = − 8 (iii) 2 9 30 25 0 x x − + = (A.B) (LHR 2017, MTN 2015) By comparing given equation with 2 ax bx c + + = 0 , we get a = = − = 9, b 30, c 25 Disc = 2 ( 30) 4(9)(25) − − = 900 – 900 = 0 (iv) 2 4 7 2 0 x x − − = (A.B) (GRW 2014, SGD 2017, MTN 2016) By comparing given equation with 2 ax bx c + + = 0 , we get 4, b 7, c 2 a = = − = − Disc = 2 b ac − 4 = 2 ( 7) 4(4)( 2) − − − = 49 + 32 = 81 Q.2 Find the nature of the roots of the following given quadratic equations and verify the result by solving the equations: Solution: (i) 2 x x − + = 23 120 0 (A.B) By comparing given equation with 2 ax bx c + + = 0 , we get a b c = = − = 1, 23, 120 Disc= 2 b ac − 4 = 2 ( 23) 4(1)(120) − − = 529 – 480 = 49 = 72 Since disc > 0 and perfect square, roots are rational (real) and unequal. Verification: 2 x x − + = 23 120 0 2 4 2 b b ac x a −  − = 2 ( 23) ( 23) 4(1)(120) 2(1) − −  − − = 2 23 7 2  = 23 7 2 x  = Either 23 7 23 7 or 2 2 x x − + = = 16 30 or 2 2 = = x x = = 8 15 Hence roots are rational and unequal. (ii) 2 2 3 7 0 x x + + = (A.B) By comparing given equation with 2 ax bx c + + = 0 , we get a b c = = = 2, 3, 7 Disc = 2 b ac − 4 = 2 (3) 4(2)(7) − = 9 – 56 = − 47
YOUTH ACADEMY YOUTUBE CHANNEL SIR USAMA Youth Academy Official Website: www.youthacademy.pk Since disc < 0, roots are complex and imaginary. Verification: 2 2 3 7 0 x x + + = 2 4 2 b b ac x a −  − = = 2 3 (3) 4(2)(7) 2(2) −  − = 3 9 56 4 −  − 3 47 4 −  − = 3 47 4 −  i = Hence roots are complex/imaginary and unequal. (iii) 2 16 24 9 0 x x − + = (A.B) By comparing given equation with 2 ax bx c + + = 0 , we get 16, b 24, c 9 a = = − = Disc = 2 b ac − 4 = 2 ( 24) 4(16)(9) − − = 576 – 576 = 0 Since disc = 0, roots are rational (real) and equal. Verification: 2 16 24 9 0 x x − + = 2 4 2 b b ac x a −  − = 2 ( 24) ( 24) 4(16)(9) 2(16) − −  − − = 24 0 32 x  = 24 0 32 x  = Either 24 0 32 x + = or 24 0 32 x − = 24 32 x = 24 32 x = Hence roots are rational and equal. (iv) 2 3 7 13 0 x x + − = (A.B) By comparing given equation with 2 ax bx c + + = 0 , we get a b c = = = − 3, 7, 13 Disc = 2 b ac − 4 2 = − − (7) 4(3)( 13) = 49 + 156 = 205 Since disc > 0, but not a perfect sq. roots are irrational and unequal. Verification: 2 3 7 13 0 x x + − = 2 4 2 b b ac x a −  − = 2 7 (7) 4(3)( 13) 2(3) −  − − = 7 49 56 6 −  + = 7 205 6 −  = Either 7 205 7 205 or 6 6 x x − − − + = = Hence roots are irrational and unequal. Q.3 For what value of k, the expression ( ) 2 2 k x k x + + + 2 1 4 is perfect square. (A.B + K.B) Solution: ( ) 2 2 k x k x + + + = 2 1 4 0 By comparing given equation with 2 ax bx c + + = 0 , we get ( ) 2 a k b k c = = + = , 2 1 , 4 Discriminant = 2 b ac − 4 ( ) ( )( ) 2 2 = + −   2 1 4 4 k k   ( ) 2 2 = + − 4 1 16 k k
YOUTH ACADEMY YOUTUBE CHANNEL SIR USAMA Youth Academy Official Website: www.youthacademy.pk ( ) 2 2 = + + − 4 2 1 16 k k k 2 2 = + + − 4 8 4 16 k k k 2 = + − 4 8 12 k k ( ) 2 = + − 4 1 2 3 k k As expression is a perfect square, the discriminant = 0 ( ) 2  + − = 4 1 2 3 0 k k 2 1 2 3 0 + − = k k 4 0  2 1 3 3 0 + − − = k k k 1 1 3 1 3 0 ( + − + = k k k ) ( ) (1 3 1 0 + − = k k )( ) Either 1 3 0 + = k or 1 0 − = k 3 1 k =− k =1 1 3 k = − Result 1 1, 3 k = − Q.4 Find the value of k, if the roots of the following equations are equal. (A.B + K.B) Solution: (i) 2 (2 1) 3 3 0 k x kx − + + = Here a k b k c = − = = 2 1, 3 , 3 Disc = 2 b ac − 4 2 = − − (3 ) 4(2 1)(3) k k 2 = − + 9 24 12 k k Since roots are equal, Disc = 0 2  − + = 9 24 12 0 k k 2 3(3 8 4) 0 k k − + = 2 3 8 4 0 k k − + = 3 0  2 3 6 2 4 0 k k k − − + = 3 ( 2) 2( 2) 0 k k k − − − = (k – 2) (3k – 2) = 0 Either k k − = − = 2 0 or 3 2 0  = = k k 2 or 3 2 2 3 k = Result: k = 2, 2 3 (ii) 2 x k x k + + + + = 2( 2) (3 4) 0 Here a b k c k = = + = + 1, 2 2 , 3 4 ( ) Disc = 2 b ac − 4 =   2 2( 2) 4(1)(3 4) k k + − + = 2 4( 2) 4(3 4) k k + − + = 2 4(k 4 4) 4(3 4) + + − + k k = 2 4 4 4 3 4   k k k + + − −   = 2 4  k k +   = + 4 ( 1) k k Since roots are equal, disc = 0 4 1 0 k k( + =) Either 4 0 1 0 k or k = + = k = 0 or k =−1 Result: 0, 1 k = − (iii) 2 (3 2) 5( 1) (2 3) 0 k x k x k + − + + + = (A.B + K.B) Here a k b k c k = + = − + = + 3 2, 5 1 , 2 3 ( ) Disc = 2 b ac − 4 =   2 − + 5( 1) k - 4 (3k+2) (2k+3) ( ) ( ) 2 2 = + + + + 25 1 – 4 6 9 4 6 k k k k ( ) ( ) 2 2 = + + − + + 25 2 1 4 6 13 6 k k k k 2 2 = + + 25 50 25 – 24 – 52 – 24 k k k k 2 = + k k – 2 1 Since roots are equal, disc = 0 2  + = k k – 2 1 0