Find 'K' if the equation kx( x-2)+6=0 has real and equal root
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here is the answer....
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=> kx (x-2) + 6 = 0
=> kx² - 2kx + 6 = 0
on comparing this with ax² + bx + c
we get.....
a = k ; b = -2k ; c = 6
For the roots to be real nd equal....
thn,
D = 0
=> b² - 4ac = 0
=> (-2k)² - 4 (k) (6) = 0
=> 4k² - 24k = 0
=> 4k ( k - 6 ) = 0
now....
4k = 0
=> k = 0/4
=> k = 0
k - 6 = 0
=> k = 6
mark me brainliest if it's helpful to uhhh.....!!!!!
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