Question

find tbe value of k such that x2-2kx+(7k-12) the equation has equal root

Answer 1

Proper Question : find the value of k such that x²-2kx+(7k-12) the equation has equal root ?


Method of Solution:


For Equal roots then it's Discriminant must be 0


•°• Discriminant = b² - 4ac


✏ b² - 4ac = 0 ,Where a = 1 , b = -2k and also c = (7k-12)

✏ (-2k)² - 4.1.(7k-12) = 0

✏ 4k² - 4(7k-12) = 0

✏ 4k² - 28k + 48 = 0

✏ 4(k² -7k + 12) = 0

✏ k² -7k +12 = 0

✏ k² -4k -3k +12 = 0

✏ k(k-4)-3(k-4) = 0

✏ (k-3)(k-4) = 0


•°• ↪ K - 3 = 0. or K-4 = 0

↪ k = 3 or 4


Hence, Required Value of k is 3 or 4.