Question

find'K'so that quad.eq.(K+1)X 2-2 (k+1)x+1=0 has equal roots?

Answer 1

Hi
Given
Equation (k+1)x sq-2 (k+1)x+1=0

For equal roots b sq =4ac
[-2 (k+1)]Sq =4(k+1).1
4 (k+1)sq=4 (k+1)
(K+1)sq=(k+1)
(K+1)=0
K=-1 Ans

Hope it helps. ...

Answer 2

Given that d=0
a=(k+1)
b=-2(k+1)
c=1
d=b^2-4ac
0={-2(k+1)}^2-4×(k+1)×1
0={-4(k^2+1+2k)}-4k-4
0=(-4k^2-4-8k)-4k-4
0=-4k^2-12k-8
0=4k^2+12k+8
0=k^2+3k+2
0=k^2+2k+k+2
0=k(k+2)+1(k+2)
0=(k+2) (k+2)
(k+2)=0. (k+1)=0
k=-2. or. k=-1