$\left( iii \right)\text{ }Given,\text{ }\left( x\text{ }\text{ }2 \right)\left( x\text{ }+\text{ }1 \right)\text{ }=\text{ }\left( x\text{ }\text{ }1 \right)\left( x\text{ }+\text{ }3 \right)$

By using the formula for ${{\left( a+b \right)}^{2~}}=\text{ }{{a}^{2}}+2ab+{{b}^{2}}$

$\Rightarrow ~{{x}^{2~}}~x~\text{ }2\text{ }=~{{x}^{2~}}+\text{ }2x~\text{ }3$

$\Rightarrow 3x~\text{ }1\text{ }=\text{ }0$

Since the above equation is not in the form of $a{{x}^{2}}~+~bx~+~c~=\text{ }0.$

Therefore, the given equation is not a quadratic equation.

$\left( iv \right)\text{ }Given,\text{ }\left( x\text{ }\text{ }3 \right)\left( 2x\text{ }+1 \right)\text{ }=\text{ }x\left( x\text{ }+\text{ }5 \right)$

By using the formula for ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$

$\Rightarrow 2{{x}^{2~}}\text{ }5x~\text{ }3\text{ }=~{{x}^{2~}}+\text{ }5x$

$\Rightarrow ~{{x}^{2~}}\text{ }10x~\text{ }3\text{ }=\text{ }0$

Since the above equation is in the form of $a{{x}^{2}}~+~bx~+~c~=\text{ }0$ .

Therefore, the given equation is quadratic equation.