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For equal root, kx(x - 2) + 6 = 0 value of k is (a) k = 6 (b) k = 3 (c) k = 2 (d) k = 8
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for equal roots value of discriminant=0
so your equation given\[kx(x-2)+6=0\\we~can ~write~\it ~as\\kx^2-2kx+6=0\]agree?
thx
now compare with \[ax^2+bx+c=0\\we'\ll ~get\\a=k,b=-2k,c=6\\so\\b^2-4ac=0\\4k^2=4.k.6\\k=6\]
np :)
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