A) \[a{{x}^{2}}+x(4a-b)+4a-2b+c=0\]
B) \[a{{x}^{2}}+x(4a-b)+4a+2b+c=0\]
C) \[a{{x}^{2}}+x(b-4a)+4a+2b+c=0\]
D) \[a{{x}^{2}}+x(b-4a)+4a-2b+c=0\]
Correct Answer: D
Solution :
We have \[\alpha +\beta =\frac{-b}{a}\]and \[\alpha \beta =\frac{c}{a}\] Now sum of the roots \[=2+\alpha +2+\beta =4-\frac{b}{a}\] and product of the roots \[=(2+\alpha )(2+\beta )\] \[=4+\frac{c}{a}-\frac{2b}{a}=\frac{4a+c-2b}{a}\] Hence the required equation is \[{{x}^{2}}-x\left( 4-\frac{b}{a} \right)+\frac{4a+c-2b}{a}=0\] or \[a{{x}^{2}}-x(4a-b)+4a+c-2b=0\] or \[a{{x}^{2}}+x(b-4a)+4a-2b+c=0\].
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