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Algebra 12 Online
OpenStudy (anonymous):

For equal root, kx(x - 2) + 6 = 0 value of k is (a) k = 6 (b) k = 3 (c) k = 2 (d) k = 8

OpenStudy (sidsiddhartha):

for equal roots value of discriminant=0

OpenStudy (sidsiddhartha):

so your equation given\[kx(x-2)+6=0\\we~can ~write~\it ~as\\kx^2-2kx+6=0\]agree?

OpenStudy (anonymous):

thx

OpenStudy (sidsiddhartha):

now compare with \[ax^2+bx+c=0\\we'\ll ~get\\a=k,b=-2k,c=6\\so\\b^2-4ac=0\\4k^2=4.k.6\\k=6\]

OpenStudy (sidsiddhartha):

np :)

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