CBSE[OUTSIDE DELHI]_X_Mathematics_2019_Set_III

The given quadratic equation is:

              kx (x–2) + 6 = 0

             kx² – 2kx + 6 = 0

        Comparing with ax² + bx + c = 0, we get

            a = k, b = – 2k and c = 6

        Since, roots are equal. So,

         D = 0 i.e., b² – 4ac = 0

                (–2k)² – 4k(6) = 0

                       4k² – 24k = 0

                        4k(k – 6) = 0

                         Either k = 0 (Neglect)

                              Or k = 6

          Therefore, the value of k is 6.

Since, 2² × 5³ × 3² × 17 = (2×5)×(2×5)×2×3²×17

                                             = 10 × 10 × 2 × 9 × 17
As there are two 10’s. So, number of zeroes in the end of number obtained by given product is two.