Question

Find that non-zero value of kfor which the quadratic equation kx2 + 1 -2(k-1)x+x2=0has equal roots Hence find the roots of the equation.

Answer 1

Answer:

Step-by-step explanation:

[1-2 (k-1)]^2-4k

1-2k+2-4k=0

3-6K=0

K=3/6

K=1/2

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 you find this answer useful then please mark as brainliest