Question

The value of k , for which the quadratic equation kx(x-2)+6=0 has two equal roots is​

Answer 1

Step-by-step explanation:

kx(x-2)+6=kx^2-2kx+6

if the equation hastwo equal roots then D=0

b^2-4ac=0

(2k)^2-4(k)(6)=0

4k^2-24k=0

4k(k-6)=0

so k =0 or 6

but in the equation value of k =0 is not possible because if k =0 then it will not be a quadratic

equation because in quadratic equation a shouldn't be equal to 0.

so the value of k =6

hope it helps

Answer 2

Answer:

$\sf\large\underline\purple{AnSwEr}$

kx(x-2)+6

kx²-2kx+6

Now here

a=k

b=-2k

c=6

Now for equal roots

d=0

b²-4ac=0

(-2k)²-4×k×6=0

4k²-24k=0

4k (k-6)=0

Now here either 4k=0 or k-6=0

K=0 or k=6

So values of k is

$\sf\large\underline\red{<u>k</u><u>=</u><u>0</u><u>,</u><u>6</u>}$