Question

find the volue of k for which the quadratic equation (k+4)xsquare +(k+1) x+1=0 has equal roots. ​

Answer 1

Answer:

Step-by-step explanation:

(k+4)x²+ ( k+1)X +1 = 0

Now , compare the equation

Ax² +Bx + C = 0

The equation has equal root so ,

B²- 4AC =0

Where B = k+1 , A = k+4 , C= 1

Now, (k+1)² - 4(k+ 4) ×1 = 0

k² +1 + 2k - 4k -16 = 0

k² - 2k -15 =0

k² - (5-3) k -15 =0

k²-5k + 3k - 15 = 0

k ( k - 5) + 3 ( k -5 ) = 0

( k - 5 ) ( k +3 ) = 0

Now, k -5 = 0 or k+ 3 = 0

k = 5 or k = -3