Question

Solve it with steps, and as fast as u can.
Plz send The pic of the answer if possible

Answer 1

In this equation
a=1
b=2(k-4)=2k-8
c=2k
D=b²-4ac
If the equation has equal roots, then D=0
b²-4ac=0
(2k-8)²-4×1×2k=0
4k²-32k+64-8k=0
4k²-40k+64=0
k²-10k+16=0
(k-2)(k-8)=0
k=2 or 8

Answer 2

for equal roots D=0
D = [2(k-4)]^2 -(4x2kx1)
i.e. 0 = [2(k-4)]^2 -(4x2kx1)
4(k-4)^2=8k
4k² + 64 -32k -8k=0
(4k² -40k+64=0)/4
k²-10k +16 =0
(k-8)(k-2)=0
i.e. k =8 or 2.



hope this helps. :)