Question

find the value of k for which each of the following equations has equal roots :- (3k+1)x2+2(k+1)x+k=0

Answer 1

( 3k + 1 )x^2 + 2 ( k + 1 )x + k = 0

here

a = 3k + 1

b = 2 ( k + 1 )

c = k

since it has equal roots

b^2 = 4ac

( 3k + 1 )^2 = 4( 2k + 2 )( k )

9k^2 + 6k + 1 = 8k^2 + 8k

k^2 - 2k + 1 = 0

k^2 - 2 ( 1 ) ( k ) + 1 = 0

( k - 1 )^2 = 0

k - 1.=0

k = 1

Answer 2

(3k+1)x²+2(k+1)x+k=0

If the equation have equal roots then -
b²-4ac=0

=>{2(k+1)²}-4(3k+1)(k)=0

=>{2(k²+1+2k)}-12k-4(k)=0

=>2k²+2+4k-12k-4k=0

=>2k²+2-12k=0

=>k²-6k+1=0

x=-b±√b²-4ac/2a

x=6k±√k²-6k+1/2