Nature of Roots Solution3





In this page nature of roots solution3 we are going to see solutions of some practice questions of the worksheet nature of roots.

(ii) 12 x² + 4 k x + 3 = 0

Since the roots are real and equal the value of ∆ = 0

a = 12   b = 4 k     c = 3

           ∆ = b² - 4 a c

          (4 k)² - 4 (12) (3) = 0

          16 k²  - 144 = 0

                16 k²  = 144

                 k² = 144/16

                 k² = 9

                 k = √9

                 k = ± 3

                 k = 3 , -3


(iii)  x² + 2 k (x - 2) + 5 = 0

      x² + 2 k x - 4 k + 5 = 0

      x² + 2 k x + (5 - 4 k)= 0

Since the roots are real and equal the value of ∆ = 0

a = 1   b = 2 k     c = 5 - 4 k

           ∆ = b² - 4 a c

          (2 k)² - 4 (1) (5 - 4 k) = 0

          4 k²  - 4 (5 - 4 k) = 0

          4 k²  - 20 + 16 k = 0

          4 k²  + 16 k  - 20 = 0

dividing the whole equation by 4, we get

             k²  + 4 k  - 5 = 0

             k²  + 5 k - 1 k  - 5 = 0           

             k(k + 5) - 1 (k  + 5) = 0           

            ( k - 1) (k + 5) = 0

               k - 1 = 0            k + 5 = 0

                k = 1                  k = -5


(iv) (k + 1) x² - 2 (k - 1) x  + 1 = 0

        (k + 1) x² - (2 k - 2) x  + 1 = 0

Since the roots are real and equal the value of ∆ = 0

a = k + 1   b = - (2 k - 2)      c = 1

           ∆ = b² - 4 a c

          (- (2 k - 2)² - 4 (k + 1) (1) = 0

         (2 k - 2)² - 4 (k + 1) (1) = 0

         (2 k)²+ (-2)² + 2 (2k) (-2) - 4 k - 4 = 0

          4 k²  - 8 k - 4 k - 4 + 4 = 0

          4 k²  - 12 k = 0

          4 k (k - 3) = 0

      4 k = 0             k - 3 = 0

        k = 0/4               k = 3

        k = 0

nature of roots solution3 nature of roots solution3