Question

if the equation (k+1)x^2 +4kx -8x + 2k =0 has real roots, find the range of values of k



Give me the correct answer no spam or else ill report you !!! im telling u im strict about my studies

Answer 1

Step-by-step explanation:

(k+1)x^2 +4kx-8x + 2k = 0

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

discriminant D = b^2 - 4ac

= [-4(k-2)]^2 - 4(k+1)(2k)

= 16k^2-64k+64-8k^2-8k

= 8k^2-72k+64 = 8(x^2-7x+8)

= 8 (k-1)(k-8)

when k=1,8 D= 0 ,it has real and equal roots

when k>8 D<0 , it has imaginary roots

when 1>k>8 D>0 ,it has real and distinct roots

therefore k€ [1,8]

Answer 2

Step-by-step explanation:

here is your answer [1 , 8] is your answer.