Question

(K-4)x^2+2(k-4)x+4=0 Find the value of K

Answer 1

Answer:

Let's assume this equation has real roots...

Therefore,

b^2 - 4ac is greater than or equal to zero..

(2k-8)^2 - 4(k - 4) >or = 0

4k^2 - 2(2k) (8) + 64 -4k + 16 >=0

4k^2 - 32k - 4k + 80 >=0

k^2 - 9k + 20 >=0.

k^2 - 5k - 4k + 20 >=0

(k - 5)(k + 4) >=0

Therefore there can be two values of k

5 and -4

Hope this helps....

Plz mark brainliest...