Question

if the equation x2-(2k+2)x+7k+1=0 has repeated real roots, find the value of k​

Answer 1

Solution :

Compare given Quadratic

equation x²-(2k+2)x+7k+1=0

by ax²+bx + c = 0 , we get

a = 1 , b = -(2k+2), c = 7k+1

Now ,

discreminant (D)=0

[ Real and equal roots ]

=> b² - 4ac = 0

=> [-(2k+2)]² - 4×1×(7k+1)=0

=> (2k+2)² - 28k-4 = 0

=>4k²+8k+4-28k-4=0

=> 4k² - 20k = 0

=> 4k(k - 5) = 0

=> 4k = 0 or k-5 = 0

=> k = 0 or k = 5

••••