Answer:
[tex]\text{Solution:}\\\text{Since }\alpha\text{ and }\beta\text{ are the roots of }2x^2+4x+5=0,\\\alpha+\beta=-\dfrac{4}{2}=-2\\\\\alpha\beta=\dfrac{5}{2}=2.5[/tex]
[tex]\text{i. Solution:}\\\text{The roots of quadratic equation are now }\alpha-1\text{ and }\beta-1\text{, so the quadratic}\\\text{equation will be: }\\x^2-\{(\alpha-1)+(\beta-1)\}x+\{(\alpha-1)(\beta-1)\}=0\\\text{or, }x^2-\{\alpha+\beta-2\}x+\{\alpha\beta-\alpha-\beta+1\}=0\\\text{or, }x^2-\{-2-2\}x+\{2.5-(\alpha+\beta)+1\}=0\\\text{or, }x^2+4x+\{2.5+2+1\}=0\\\text{or, }x^2+4x+5.5=0\\\text{Multiplying `10' on both sides,}\\10x^2+40x+55=0\text{ is the required equation.}[/tex]
[tex]\text{ii. Solution:}\\\text{The roots of quadratic equation are }\alpha^2\text{ and }\beta^2,\text{ so the quadratic equation}\\\text{will be:}\\x^2-(\alpha^2+\beta^2)x+\alpha^2\beta^2=0\\\text{or, }x^2-\{(\alpha+\beta)^2-2\alpha\beta\}x+(\alpha\beta)^2=0\\\text{or, }x^2-\{(-2)^2-2(2.5)\}x+(2.5)^2=0\\\text{or, }x^2-\{4-5\}x+6.25=0\\\text{or, }x^2+x+6.25=0\\\text{Multiplying both sides by 4, }\\4x^2+4x+25=0\text{ is the required quadratic equation.}[/tex]
[tex]\text{iii. Solution:}\\\text{The roots of the equation are now }\dfrac{1}{\alpha}\text{ and }\dfrac{1}{\beta},\text{ so the quadratic equation will}\\\text{be:}\\x^2-\biggl\{\dfrac{1}{\alpha}+\dfrac{1}{\beta}\biggl\}x+\bigg(\dfrac{1}{\alpha}.\dfrac{1}{\beta}\bigg)=0[/tex]
[tex]\text{or, }x^2-\bigg(\dfrac{\beta+\alpha}{\alpha\beta}\bigg)x+\bigg(\dfrac{1}{\alpha\beta}\bigg)=0[/tex]
[tex]\text{or, }x^2-\bigg(\dfrac{-2}{2.5}\bigg)x+\dfrac{1}{2.5}=0\\\\\text{or, }x^2+\dfrac{2}{2.5}x+4=0\\\\[/tex]
[tex]\text{Multiplying by 5 on both sides,}\\\\5x^2-5\times\dfrac{2}{2.5}x+\dfrac{5}{2.5}=0\\\\\text{or, }5x^2+4x+2=0\text{ is the required quadratic equation.}[/tex]
