Question

find the value of k so that the equation (k+4)x²+(k+1)x+1=0 has real and equal roots​

Answer 1

Answer:

✅Here's your answer

If the solution has real and equal roots than

D=0

As

D= (k+1) ^2 - 4×(k+4) ×1

D= k^2 +2k+1-4k-16

D= k^2-2k-15

So D=0

k^2-2k-15=0

k(k-2)-15 =0

k-2=15

k=17...

Hope it helps✌✌...