Question

# If  $$\alpha$$ and $$\beta$$ are the roots of the equation $${x^2} - 7x + 1 = 0$$ then find the value $$\frac{1}{{{{(\alpha - 7)}^2}}} + \frac{1}{{{{(\beta - 7)}^2}}}$$

Solution

## $$\Rightarrow$$   $$\alpha$$ and $$\beta$$ are the roots of the equation $$x^2-7x+1=0$$Here, $$a=1,\,b=-7,\,c=1$$$$\Rightarrow$$  $$\alpha\beta=\dfrac{c}{a}=\dfrac{1}{1}=1$$        ----- ( 1 )$$\Rightarrow$$  $$\alpha^2\beta^2=1$$             ----- ( 2 )$$\Rightarrow$$  $$\alpha+\beta=\dfrac{-b}{a}=-\dfrac{-7}{1}=7$$         ----- ( 3 )$$\Rightarrow$$  $$(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta$$$$\Rightarrow$$  $$(7)^2=\alpha^2+\beta^2+2(1)$$          [ Using ( 1 ) and ( 3 ) ]$$\Rightarrow$$  $$49=\alpha^2+\beta^2+2$$$$\therefore$$  $$\alpha^2+\beta^2=47$$                  ------- ( 4 )Now,$$\Rightarrow$$  $$\dfrac{1}{(\alpha-7)^2}+\dfrac{1}{(\beta-7)^2}=\dfrac{(\beta-7)^2+(\alpha-7)^2}{(\alpha-7)^2(\beta-7)^2}$$                                              $$=\dfrac{\beta^2-14\beta+49+\alpha^2-14\alpha+49}{(\beta^2-14\beta+49)(\alpha^2-14\alpha+49)}$$                                              $$=\dfrac{\alpha^2+\beta^2-14(\alpha+\beta)+98}{\alpha^2\beta^2-14\alpha\beta^2+49\beta^2-14\alpha^2\beta+196\alpha\beta-686\beta+49\alpha^2-686\alpha+2401}$$                                              $$=\dfrac{47-14(7)+98}{\alpha^2\beta^2-14\alpha\beta(\alpha+\beta)+49(\alpha^2+\beta^2)+196\alpha\beta-686(\alpha+\beta)+2401}$$              [ Using ( 3 ) and ( 4 ) ]                                              $$=\dfrac{47}{1-14(1)(7)+49(47)+196(1)-686(7)+2401}$$                                              $$=\dfrac{47}{1-98+2303+196-4802+2401}$$                                              $$=\dfrac{47}{1}$$$$\therefore$$   $$\dfrac{1}{(\alpha-7)^2}+\dfrac{1}{(\beta-7)^2}=47$$Mathematics

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