If the equation

Solution:

for $2 x^{2}-k x+x+8=0$

given; roots one real and equal

i.e. $D=0$   (Discriminat)

i. e. $b^{2}-4 a c=0$

Here $b=-k+1$

$a=2$

$c=8$

with reference to standard equations

$a x^{2}+b x+c=0$

$\Rightarrow(1-K)^{2}-4(2) 8=0$

i.e. $K^{2}+1-2 K-64=0$

i.e. $K^{2}-2 K-63=0$

i. e. $K^{2}-9 K+7 K-63=0$

i.e. $(K-9)(k+7)=0$

i.e. $K=9$ or $-7$