Question

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

Answer 1

$\huge\bf\mathfrak\green{SOLUTION}$

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

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

if u didn't understand this than 2 way is:-

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___