Solutions:

$\left( i \right)\text{ }2{{x}^{2}}~+~kx~+\text{ }3\text{ }=\text{ }0$

Comparing the given equation with $a{{x}^{2}}~+~bx~+~c~=\text{ }0$ , we get,

$a~=\text{ }2,~b~=\text{ }k\text{ }and~c~=\text{ }3$

As we know, Discriminant $=~{{b}^{2}}~\text{ }4ac$

$=\text{ }{{\left( k \right)}^{2}}~\text{ }4\left( 2 \right)\text{ }\left( 3 \right)$

$=~{{k}^{2}}~\text{ }24$

For equal roots, we know,

Discriminant = 0

${{k}^{2}}~\text{ }24\text{ }=\text{ }0$

${{k}^{2}}~=\text{ }24$

$k\text{ }=\text{ }\pm \surd 24~=\text{ }\pm 2\surd 6$

$\left( ii \right)~kx\left( x~\text{ }2 \right)\text{ }+\text{ }6\text{ }=\text{ }0$

$or~k{{x}^{2}}~\text{ }2kx~+\text{ }6\text{ }=\text{ }0$

Comparing the given equation with $a{{x}^{2}}~+~bx~+~c~=\text{ }0$ , we get

$a~=~k,~b~=\text{ }\text{ }2k~and~c~=\text{ }6$

We know, Discriminant $=~{{b}^{2}}~\text{ }4ac$

$=\text{ }{{\left( \text{ }\text{ }2k \right)}^{2}}~\text{ }4\text{ }\left( k \right)\text{ }\left( 6 \right)$

$=\text{ }4{{k}^{2}}~\text{ }24k$

For equal roots, we know,

${{b}^{2}}~\text{ }4ac~=\text{ }0$

$4{{k}^{2}}~\text{ }24k~=\text{ }0$

$4k~\left( k~\text{ }6 \right)\text{ }=\text{ }0$

$Either\text{ }4k~=\text{ }0\text{ }or~k~=\text{ }6\text{ }=\text{ }0$

$k~=\text{ }0\text{ }or~k~=\text{ }6$

However, if k = 0, then the equation will not have the terms ‘x²‘ and ‘x‘.

Therefore, if this equation has two equal roots, k should be 6 only