Question

2. For what value of k, the given equation has real
and equal roots(k + 1) x2 – 2 (k – 1) x + 1=0​

Answer 1

Step-by-step explanation:

★ For roots to be real D = 0 where D = b²- 4ac

=> we have, a = k+1 , b = -2(k+1) and c = 1

=> substituting these values in D

=> b²- 4ac = 0 => {-2(k+1)}² - 4(k+1)(1)

=> 4(k²+2k +1) -4(k+1) = 0

=> 4[ k²+2k+1-k-1] = 0

=> k²+k = 0/4

=> k²+k = 0

=> k(k+1) = 0

=> k =0 OR k+1 = 0

=> k = -1,0

So For K = -1 , 0 given equation has real and equal roots.

Hope It helps you Buddy!!!........ :) :)