Question

for what values of k the equation 4x^-2(k+1)x+(k+1)=0

Answer 1

Here, a=4 , b=-2(k+1) , c=(k+1)
D = b^2-4ac
= {-2(k+1)}^2 -4×4×(k+1)
= {4(k+1)^2} -16(k+1)
= {4(k^2+1+2k)} -16(k+1)
= 4k^2+8k+4-16k-16
= 4k^2-8k-12
= 4(k^2-2k-3) =0
= k^2-2k-3=0
= k^2-2k =3
= k(k-2) =3
= k=3 or k-2=3, k=5