Question

5. Find the value(s) of k for which the roots are real and equal in the equation (k+4)x² + (k+1)x + 1 = 0​

Answer 1

Step-by-step explanation:

(k+4)x2 +(k+1)x+1=0 D=b2 -4ac =(k+1)2-4(k+4)(1) =k2 +2k+1-4k-16 =k2 -2k-15 For equal roots, D=0 D=0 K2 -2k-15=0 k2 -5k+3k-15=0 k(k-5)+3(k-5)=0 (k+3)(k-5)=0 k+3=0 OR k-5=0 k = -3, k = 5Read more on Sarthaks.com - https://www.sarthaks.com/365876/find-the-value-of-k-for-which-the-quadratic-equation-k-4-x-2-k-1-x-1-0-has-equal-roots

Answer 2

Answer:

If roots are equal than

D=0

b²-4ac=0

(k+1)²-4×k+4×1 =0

k²+1+2k-4k-16=0

k²-2k-15=0

now using splitting the middle term

k²-5k+3k-15=0

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

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

k=5,-3

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