Question

For what value of k the quadratic equation (k+4)x^2+(k+4)x-1=0

Answer 1

For equal roots it must have D=0 where D= b^2-4ac (b square minus 4ac)

or, b^2-4ac=0

or, (k+1)^2-4(k+4)(1)=0

or, k^2+2k+1-4k-4=0

or, k^2-2k-3=0

or, k^2-3k+k-3=0

or, k(k-3)+1(k-3)=0

or, k-3=0 and k+1=0

Hence the value of k should be 3 or -1.

Hope it helped.