\\(\\begin{array}{l}{\ ext { (i) Let the breadth of the plot be } x \\mathrm{m} \ ext { . }} \\\\ {\ ext { Hence, the length of the plot is }(2 x+1) \\mathrm{m} \ ext { . }} \\\\ {\ ext { Area of a rectangle = Length } \ imes \ ext { Breadth }} \\\\ {\ herefore 528=x(2 x+1)} \\\\ {\\Rightarrow 2 x^{2}+x-528=0}\\end{array}\\)
\\(\\begin{array}{l}{\ ext { (ii) Let the consecutive integers be } x \ ext { and } x+1} \\\\ {\ ext { It is given that their product is } 306 \ ext { . }} \\\\ {\ herefore x(x+1)=306 \\Rightarrow x^{2}+x-306=0}\\end{array}\\)
\\(\\begin{array}{l}{\ ext { (iii) Let Rohan's age be } x \ ext { . }} \\\\ {\ ext { Hence, his mother's age }=x+26} \\\\ {3 \ ext { years hence, }} \\\\ {\ ext { Rohan's age }=x+3} \\\\ {\ ext { Mother's age }=x+26+3=x+29} \\\\ {\ ext { It is given that the product of their ages after } 3 \ ext { years is } 360 \ ext { . }} \\\\ {\ herefore(x+3)(x+29)=360} \\\\ {\\Rightarrow x^{2}+32 x-273=0}\\end{array}\\)
\\(\\begin{array}{l}{\ ext { (iv) Let the speed of train be } x \\mathrm{km} / \\mathrm{h} \ ext { . }} \\\\ {\ ext { Time taken to travel } 480 \\mathrm{km}=\\frac{480}{x} \ ext { hrs }} \\\\ {\ ext { In second condition, let the speed of train }=(x-8) \\mathrm{km} / \\mathrm{h}} \\\\ {\ ext { It is also given that the train will take } 3 \ ext { hours to cover the same }} \\\\ {\ ext { distance. }} \\\\ {\ ext { Therefore, time taken to travel } 480 \\mathrm{km}=\\left(\\frac{480}{x}+3\\right)_{\ ext { hrs }}}\\end{array}\\)
\\(\\begin{array}{l}{\ ext { Speed } \ imes \ ext { Time }=\ ext { Distance }} \\\\ {(x-8)\\left(\\frac{480}{x}+3\\right)=480} \\\\ {\\Rightarrow 480+3 x-\\frac{3840}{x}-24=480} \\\\ {\\Rightarrow 3 x-\\frac{3840}{x}=24} \\\\ {\\Rightarrow 3 x^{2}-24 x+3840=0} \\\\ {\\Rightarrow x^{2}-8 x+1280=0}\\end{array}\\)