Question

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

Answer 1

given: kx2 + 1-2(k-1)x + x2
sol;
=kx2 + x2 -2(k-1) +1
x2 (k+1) - 2(k-1) +1
here,
a=(k+1)   b= -2(k-1) & c= 1
for real and equal roots;
D=0
=b2- 4ac=0

Putting the values of a.b and c

[-2(k-1)]2 - 4 (k+1)(1)=0

then on opening the brackets we'll get;
(4k)(k-3)=0
=   4k=0             and        (k-3)=0
=     k=0               or           k=3

Answer 2

Hey mate!!
Here's ur answer:

b^2 -4ac = 0
(-2k+2)^2 - 4(k+1) =0
4k^2 -8k +4 -4k -4 =0
4k^2 -12k=0
4k^2 = 12k
___k =3___


Hope u understand!


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