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

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

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.

   Reply