Question

Find the value of k so that the equation (3k+ 1 )x2 +2 (k+1 )x + k = 0 may have equal roots

Answer 1

3k+1)x^2+2(k+1)x+1=0 
For equal roots, D=0 
b^2-4ac=0 [2(k+1)]^2-4(3k+1)×1=0 
4(k^2+1+2k)-12k-4=0 
4(k^2+1+2k-3k-1)=0 
k^2+1+2k-3k-1=0 
k^2-k=0 
k(k-1)=0 
k=0 or k-1=0 
(rejected as k can't be 0) 
=>k-1=0 
k=1

Answer 2

3k+1)x^2+2(k+1)x+1=0 

For equal roots, D=0 

b^2-4ac=0 [2(k+1)]^2-4(3k+1)×1=0 

4(k^2+1+2k)-12k-4=0 

4(k^2+1+2k-3k-1)=0 

k^2+1+2k-3k-1=0 

k^2-k=0 

k(k-1)=0 

k=0 or k-1=0 

(rejected as k can't be 0) 

=>k-1=0 

k=1

Read more on Brainly.in - https://brainly.in/question/5695650#readmore