Question

Find the value of k for which the quadratic equation (k+1)x2 -6(k+1) x +3(k+9) =0 has equal roots . Also find the roots

Answer 1

Answer:

For equal roots, D=0

[-6(k+1)]^2-4(k+1).3(k+9)=0

=> 36k2+72k+36-12(k2+10k+9)=0

=> 36k2+72k+36-12k2-120k-108=0

=> 24k2-48k-72=0

=> k2-2k-3=0

=> (k-3)(k+1)=0

=> k = 3 or - 1

Clearly k can't be - 1. So, k = 3