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Answers & Comments
UnknownDude
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
Answers & Comments
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
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. :)