Find k if (2k+1)x^2-(7k+2)x+6k+1=0, has equal zeroes
Answers
Answered by
1
Comparing with Axx + Bx + C = 0,
A = 2k + 1, B = - ( 7k + 2 ), C = 6k + 1
Since equal zeros, discrimant D = 0
D = B*B - 4*A*C
D = (-(7k+2))*(-(7k+2)) - 4*(2k+1)*(6k+1)
D = (49k*k + 4 + 28k) - (48k*k + 32k + 4)
D = k*k - 4k = k(k-4) = 0
Hence k = 0 and k = 4
A = 2k + 1, B = - ( 7k + 2 ), C = 6k + 1
Since equal zeros, discrimant D = 0
D = B*B - 4*A*C
D = (-(7k+2))*(-(7k+2)) - 4*(2k+1)*(6k+1)
D = (49k*k + 4 + 28k) - (48k*k + 32k + 4)
D = k*k - 4k = k(k-4) = 0
Hence k = 0 and k = 4
shuchidatta29:
I think it should be (49k*k+4+28k)-(48k*k+28k+4).
But answer is still correct
Thnc
Similar questions