Question

If the quadratic equation (k+1)x2-2(k-1)x+1=0 have Real and equal roots then, find the value of k

Answer 1

Answer:

Step-by-step explanation:

Answer 2

(k+1)x² - 2(k+1)x + 1 = 0

Comparing it in the form

ax² + bx + c = 0

a = k+1

b = -2(k+1)

c = 1

For equal roots: b² - 4ac = 0

=> {-2(k+1)}² - 4(k+1) (1) = 0

=> {-2k - 2}² - 4k - 4 = 0

=> (-2k)² + (2)² - 2(-2k) (2) - 4k - 4 = 0

=> 4k² + 4 + 8k - 4k - 4 = 0

=> 4k² + 4k = 0

=> 4k(k+1) = 0

=> 4k = 0 ; k + 1 = 0

=> k = 0 ; k = -1

Putting , k = -1 :

(-1 +1)x² - 2(-1 +1)x + 1 = 0

=> 0x² -2(0)x + 1 = 0 (which is not possible)

So, $\fbox{ k = 0 }$