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The Brainliest Answer!
2016-01-16T13:53:30+05:30
since the eqn has equal roots,we get
b²-4ac=0
[-2(k-1)]² - 4(k+1)(1)=0
(-2k+2)² -4(k+1)=0
4k² - 8k +4 -4k -4 =0
4k² -12k =0
4k² =12k
k=12k/4k
k=3
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2016-01-16T14:10:59+05:30
Given equation : (k+1) x² - 2(k-1)x +1 = 0
When compared to the general form
ax² + bx +c = 0
a =(k+1) , b = -2(k-1) , c =1
Given that the zeroes of the polynomial are equal then by 
Discriminant(Δ) = 0 
⇒ b² - 4ac = 0
⇒ [-2(k-1)]² - 4(k+1)(1) = 0
⇒ (-2k + 2)² -4k+4 = 0
⇒  4k² -8k - 4 - 4k +4 = 0
⇒  4k² -12k = 0
⇒  4k(k -3) = 0
⇒   k-3 = 0/4k
⇒   k-3 = 0
∴    k =3
 
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