Question

Find the value of k if the rootsof quadratic equations kx(x-3) +6=0

Answer 1

kx(x-3)=0

=>

$kx^{2} - 3kx = 0$

so,

${kx}^{2} - 3k - 0$

here, a= k

b= 3k

c= 0

${b}^{2} - 4ac$

${3k}^{2} - 4(k)(0) \\ {3k}^{2} - 0 \\ {3k}^{2} = 0 \\ {k}^{2} = 0 \\ k = 0$