Question

find the value of k dor which the quadratic equation (k+4)x2+(k+1)x+1=0 as equal roots​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Here, a= 4-k, b= 2k+4, c= 8k+1

The discriminate (D) = b2 – 4ac

= (2k+4)2 – 4(4-k)(8k+1)

= (4k2+16+ 16k) -4(32k+4-8k2-k)

= 4(k2 +8k2+4k-31k+4-4)

=4(9k2-27k)

D = 4(9k2-27k)

The given equation is a perfect square

D= 0

4(9k2-27k) = 0

9k2-27k=0

Taking out common of of 3 from both sides and cross multiplying

= k2 -3k =0

= K (k-3) =0

Either k=0

Or k =3

The value of k is to be 0 or 3 in order to be a perfect square.