Question

Find the values of k for which the given equation has real and equal roots:(k+1)x2−2(k−1)x+1=0​

Answer 1

Answer:

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Answer 2

Answer:

$Given \: equation \: \: is  \: (k+1) \: x2 \: −2 \: (k−1) \: x+1=0</p><p></p><p>$

Step-by-step explanation:

$⟹ \: Discriminant =0$

⟹[−2(k−1)]

⟹[−2(k−1)] 2

⟹[−2(k−1)] 2 −4(k+1)(1)=0

⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1)

⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2

⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)

⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)⟹k

⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)⟹k 2

⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)⟹k 2 −2k+1=k+1

⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)⟹k 2 −2k+1=k+1⟹k

⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)⟹k 2 −2k+1=k+1⟹k 2

⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)⟹k 2 −2k+1=k+1⟹k 2 =3k

⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)⟹k 2 −2k+1=k+1⟹k 2 =3k⟹k=0 or 3