Solution:

Given,

\[{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{2}\left( \mathbf{k}\text{ }+\text{ }\mathbf{1} \right)\mathbf{x}\text{ }+\text{ }{{\mathbf{k}}^{\mathbf{2}}}~=\text{ }\mathbf{0}\]

It’s of the form of  \[a{{x}^{2~}}+\text{ }bx\text{ }+\text{ }c\text{ }=\text{ }0\]

a =$1$, b = $-2(k + 1)$, c =  

For the quadratic equation to have real roots \[D\text{ }=\text{ }{{b}^{2~}}\text{ }4ac\text{ }=\text{ }0\]

\[D\text{ }=\text{ }{{\left( -2\left( k\text{ }+\text{ }1 \right) \right)}^{2}}~\text{ }4\left( 1 \right)\left( {{k}^{2}} \right)\text{ }=\text{ }0\]

\[\Rightarrow 4{{k}^{2}}~+\text{ }8k\text{ }+\text{ }4\text{ }\text{ }4{{k}^{2}}~=\text{ }0\]

⇒ $8k + 4 = 0$

⇒ $k = -4/8$

⇒ $k = -1/2$

The value of k should $-1/2$ to have real and equal roots.

Solution:

Given :

\[{{k}^{2}}{{x}^{2}}~\text{ }2\text{ }\left( 2k\text{ }\text{ }1 \right)x\text{ }+\text{ }4\text{ }=\text{ }0\]   

It’s of the form of \[a{{x}^{2~}}+\text{ }bx\text{ }+\text{ }c\text{ }=\text{ }0\]

\[a\text{ }=\text{ }{{k}^{2}}\], b = $–2 (2k – 1)$, c = $4$

For the quadratic equation to have real roots \[D\text{ }=\text{ }{{b}^{2~}}\text{ }4ac\text{ }=\text{ }0\]

\[D\text{ }=\text{ }{{\left( -2\left( k\text{ }+\text{ }1 \right) \right)}^{2}}~\text{ }4\left( 4 \right)\left( {{k}^{2}} \right)\text{ }=\text{ }0\]

\[\Rightarrow 4{{k}^{2}}~+\text{ 4}k\text{ }+\text{ 1 }\text{ }4{{k}^{2}}~=\text{ }0\] (dividing by 4 both sides)

⇒ $-4k + 1 = 0$

⇒ $k = 1/4$

The value of k should $1/4$ to have real and equal roots.