Question

Find that non zero value of k, for which the quadratic equation kx^2+x^2-2(k-1)x+1=0 has equal roots. Hence find the root of the equation ​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

Given equation is (k+1)x

Given equation is (k+1)x 2

Given equation is (k+1)x 2 −2(k−1)x+1=0

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots⟹Discriminant =0

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots⟹Discriminant =0⟹[−2(k−1)]

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots⟹Discriminant =0⟹[−2(k−1)] 2

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots⟹Discriminant =0⟹[−2(k−1)] 2 −4(k+1)(1)=0

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots⟹Discriminant =0⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1)

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots⟹Discriminant =0⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots⟹Discriminant =0⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots⟹Discriminant =0⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)⟹k

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots⟹Discriminant =0⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)⟹k 2

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots⟹Discriminant =0⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)⟹k 2 −2k+1=k+1

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots⟹Discriminant =0⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)⟹k 2 −2k+1=k+1⟹k

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots⟹Discriminant =0⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)⟹k 2 −2k+1=k+1⟹k 2

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots⟹Discriminant =0⟹[−2(k−1)] 2 −4(k+1)(1)=0⟹(k−1) 2 =(k+1)⟹k 2 −2k+1=k+1⟹k 2 =3k

Given equation is (k+1)x 2 −2(k−1)x+1=0This equation has real and equal roots⟹Discriminant =0⟹[−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