Question

find the value of K for which the given equation has equal roots (k-12) x^2+2(k-12)x+2=0​

Answer 1

Answer:

The values of k for which the given equation has equal roots are 14 and 12.

Step-by-step explanation:

Given polynomial is;

              (k-12) x^2+2(k-12)x+2=0;

here,  a = (k-12);

         b = -2(k-12);

          c = 2

For equal roots, we know a discriminant and that is;

               b² - 4ac = 0;

Thus , (-2(k-12))² - 4(k-12)(2) = 0;

          4(k-12)²- 4(k-12)(2) = 0;

First value of k will be;

        4(k-12){(k-12) - 2} = 0;

                  (k-12) - 2 = 0 / 4(k-12);

                   k - 12 - 2 = 0;

                  k - 14= 0;

                  k = 14.

Second value of k will be;

        4(k-12){(k-12) - 2} = 0;

        4(k-12){(k-12) - 2} = 0 / {(k-12) - 2};

                   4(k-12) = 0;

                  k - 12= 0;

                  k = 12.

That's all.