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## Answers

The Brainliest Answer!

**since the eqn has equal roots,we get**

b²-4ac=0

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**

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