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