Question

Answer it.....................

Answer 1

Answer:

Option B is correct (8,2)

$\red{ \huge \mathfrak{solution}}$

Here

a=1

b=2(k-4)

and

c=2k

for equal roots

D=0

=> b^2-4ac=0

$\rightarrow \: {(2(k - 4))}^{2} - 4(1)(2k) = 0 \\ \\ \rightarrow \: 4 {k}^{2} - 40k + 64 = 0 \\ \\ \rightarrow \: {k}^{2} - 10k + 16 = 0 \\ \\ \rightarrow \: (k - 2)(k - 8) = 0 \\ \\ \rightarrow \: k = \: 2 \: and \: 8$

soln also is in attachment

Answer 2

Answer:

Option B is correct

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