Question

find the value of k if the equation has two equal roots: kx (x-2) + 6=0

Answer 1

The given equation can be written in the form of
k {x}^{2} - 2kx + 6 = 0kx2−2kx+6=0 

Now it has equal roots that means the Discriminant is 0
D = 0

Also
b2−4ac=04k2−4(k)(6)=04k2−24k=04k(k−6)=0k−6=0k=6​ 

Hence the value of K is 6.

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Answer 2

kx (x-2) + 6 = 0

(k)x^2 + (-2k)x + 6 = 0
(a)x^2 + (b)x + c = 0. (General form)

For equal roots,
D = 0

b^2 - 4ac = 0. (General form)
(-2k)^2 - 4(k)(6) = 0
4k^2 - 24k = 0
k^2 - 6k = 0
k (k-6) = 0

k = 6, 0

=> 0 is not possible because the resultant equation will not remain quadratic

=> k = 6

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