Question

x^2-4x-1=2k(x-5) where k is constant,has two real roots​

Answer 1

Step-by-step explanation:

x^2-4x-2kx-1+10k=0

x^2-(2k+4)+10k-1=0

for two real roots

b^2-4ac>0

(2k+4)^2-4(10k-1)>0

4k^2+16k+16-40k+4>0

4k^2-24k+20>0

k^2-6k+5>0

(k-1)(k-5)>0

k<1 or k>5