Question

find the value of K if the equation Kx(x-2)+6=0 has real and equal roots?

plz do give with explanation​

Answer 1

Answer:

$0 or \frac{3}{2}$

Step-by-step explanation:

A quadratic equation of the form

$ax^{2} +bx+c=0$

has real and equal roots if

$b^{2} -4ac=0$

On expanding Kx(x-2)+6=0

$Kx^{2} -2Kx+6=0$

a= K,b=-2K and c= 6

if $b^{2} - 4ac = 0$ then

$(-2K)^{2}-4*K*-K*6=0\\4K^{2} -6K=0\\K=0 \\or K=\frac{3}{2}$