Find the value of k for which the equation kx2 _kx+1=0
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kx2 - kx + 1 = 0
D = b2 - 4ac ( a = k, b = - k, c = 1 )
= ( - k ) square - 4 (k)(1)
= k square - 4k
= k ( k - 4 )
k = 0 , k = +4
"0" cannot be the value.
therefore the value of k is + 4.
hope it will be helpful.....
please mark as brainliest.
D = b2 - 4ac ( a = k, b = - k, c = 1 )
= ( - k ) square - 4 (k)(1)
= k square - 4k
= k ( k - 4 )
k = 0 , k = +4
"0" cannot be the value.
therefore the value of k is + 4.
hope it will be helpful.....
please mark as brainliest.
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