Question

find the value of 'k' for quadratic equation kx^+(2k+4)x+9​

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