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Hint: Here, we will be putting the discriminant of the given quadratic equation equal to zero in order for this equation to have two equal roots. From here we will obtain the value of unknown k.

Complete step-by-step answer:

The given quadratic equation is \[

kx\left( {x - 2} \right) + 6 = 0 \\

\Rightarrow k{x^2} - 2kx + 6 = 0{\text{ }} \to {\text{(1)}} \\

\]

As we know that for any general quadratic equation \[a{x^2} + bx + c = 0{\text{ }} \to {\text{(2)}}\] to have two equal roots, the discriminant should be equal to zero where the discriminant is given by

${\text{d}} = \sqrt {{b^2} - 4ac} \to {\text{(3)}}$

By comparing equations (1) and (2), we get

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

It is also given that the given quadratic equation given by equation (1) has two equal roots.

So, d=0 for the given quadratic equation.

$

\Rightarrow \sqrt {{{\left( { - 2k} \right)}^2} - 4\left( k \right)\left( 6 \right)} = 0 \\

\Rightarrow \sqrt {4{k^2} - 24k} = 0 \\

\Rightarrow 4{k^2} - 24k = 0 \\

\Rightarrow 4k\left( {k - 6} \right) = 0 \\

$

Either $

4k = 0 \\

\Rightarrow k = 0 \\

$ or $

\left( {k - 6} \right) = 0 \\

\Rightarrow k = 6 \\

$

But here k=0 is neglected because if k=0 the given quadratic equation $kx\left( {x - 2} \right) + 6 = 0$ reduces to 6=0 which is not true.

So, k=6 is the only value for which the given quadratic equation has two equal roots.

Note: In these types of problems, the values of the roots corresponding to the any quadratic equation \[a{x^2} + bx + c = 0\] are given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. If the quadratic equation has two equal roots then the value of that root is given by $x = \dfrac{{ - b}}{{2a}}$.

Complete step-by-step answer:

The given quadratic equation is \[

kx\left( {x - 2} \right) + 6 = 0 \\

\Rightarrow k{x^2} - 2kx + 6 = 0{\text{ }} \to {\text{(1)}} \\

\]

As we know that for any general quadratic equation \[a{x^2} + bx + c = 0{\text{ }} \to {\text{(2)}}\] to have two equal roots, the discriminant should be equal to zero where the discriminant is given by

${\text{d}} = \sqrt {{b^2} - 4ac} \to {\text{(3)}}$

By comparing equations (1) and (2), we get

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

It is also given that the given quadratic equation given by equation (1) has two equal roots.

So, d=0 for the given quadratic equation.

$

\Rightarrow \sqrt {{{\left( { - 2k} \right)}^2} - 4\left( k \right)\left( 6 \right)} = 0 \\

\Rightarrow \sqrt {4{k^2} - 24k} = 0 \\

\Rightarrow 4{k^2} - 24k = 0 \\

\Rightarrow 4k\left( {k - 6} \right) = 0 \\

$

Either $

4k = 0 \\

\Rightarrow k = 0 \\

$ or $

\left( {k - 6} \right) = 0 \\

\Rightarrow k = 6 \\

$

But here k=0 is neglected because if k=0 the given quadratic equation $kx\left( {x - 2} \right) + 6 = 0$ reduces to 6=0 which is not true.

So, k=6 is the only value for which the given quadratic equation has two equal roots.

Note: In these types of problems, the values of the roots corresponding to the any quadratic equation \[a{x^2} + bx + c = 0\] are given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. If the quadratic equation has two equal roots then the value of that root is given by $x = \dfrac{{ - b}}{{2a}}$.

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