Question

find the value of k for which the equation (k-5)n^2+2(k-5)n+2=0 has equal roots​

Answer 1

Step-by-step explanation:

(k-5)n²+2(k-5)n+2=0

(k-5)n²+(2k-10)n+2=0

Comparing with the form an² + bn +c= 0

Here, a= (k-5)

b= (2k-10) ; c = 2

For equal roots:

b² - 4ac = 0

=>(2k-10)² - 4(k-5)2 = 0

=> 4k² + 100 - 40k - 8k + 40 = 0

=> 4k² - 48k + 140 = 0

=> k² - 12k + 35 = 0

=> k² - 7k - 5k + 35 = 0

=> k(k - 7) - 5(k - 7) = 0

=> (k - 5) (k - 7) = 0

=> k = 5, 7