Question

Please solve this!!!!!!!!

Answer 1

Hey mate, I don't know if it is correct but have tried my best.......

x= 0

(k-2)(0)^2 - (5+k)(0) + 16 = 0

(k+k) - (5-2) + 16 = 0

2k - 3 + 16 = 0

2k - 3 = 16

k - 3 = 16/2

k - 3 = 8

k = 8 + 3

k = 11

Hope it was helpful...

Answer 2

For real and equal root of equation

$b^2-4ac=0\\\\(5+k)^2-4\times16\times(k-2)=0\\\\(5+k)^2=4\times16\times(k-2)\\25+10k+k^2=64k-128\\k^2-54k+153=0\\(k-51)(k-3)=0\\k=51,3$

so the value of k is 51,3