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Question-214408

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Question Number 214408 by Abdullahrussell last updated on 07/Dec/24
Answered by A5T last updated on 07/Dec/24
Let f(x)=x^4 −10x^3 +37x^2 −60x+32 f(1)=f(4)=0⇒f(x)=(x^2 −5x+4)(x^2 −5x+8) ⇒Σ(1/(a^2 −5a+10))=(1/6)+(1/6)+(1/2)+(1/2)=1(1/3)=(4/3)
$${Let}\:{f}\left({x}\right)={x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{3}} +\mathrm{37}{x}^{\mathrm{2}} −\mathrm{60}{x}+\mathrm{32} \\ $$$${f}\left(\mathrm{1}\right)={f}\left(\mathrm{4}\right)=\mathrm{0}\Rightarrow{f}\left({x}\right)=\left({x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}\right)\left({x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{8}\right) \\ $$$$\Rightarrow\Sigma\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{5}{a}+\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{1}\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$
Commented by Abdullahrussell last updated on 08/Dec/24
Sir, kindly details
$$\:{Sir},\:{kindly}\:{details} \\ $$
Answered by mr W last updated on 08/Dec/24
say x^2 −5x+10=t t^2 =(x^2 −5x+10)^2 =x^4 +25x^2 +100−10x^3 +20x^2 −100x =x^4 −10x^3 +37x^2 −60x+32+8x^2 −40x+68 =8x^2 −40x+68 =8(x^2 −5x+10)−12 =8t−12 t^2 −8t+12=0 (t−2)(t−6)=0 t_(1,2) =2, 6 ⇒(1/t_(1,2) )=(1/2), (1/6) ⇒(1/(x_(1,2) ^2 −5x_(1,2) +10))=(1/2) ⇒(1/(x_(3,4) ^2 −5x_(3,4) +10))=(1/6) i.e. (1/(a^2 −5a+10))+(1/(b^2 −5b+10))+(1/(c^2 −5c+10))+(1/(d^2 −5d+10)) =(1/2)+(1/2)+(1/6)+(1/6) =(4/3) ✓
$${say}\:{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{10}={t} \\ $$$${t}^{\mathrm{2}} =\left({x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{10}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:={x}^{\mathrm{4}} +\mathrm{25}{x}^{\mathrm{2}} +\mathrm{100}−\mathrm{10}{x}^{\mathrm{3}} +\mathrm{20}{x}^{\mathrm{2}} −\mathrm{100}{x} \\ $$$$\:\:\:\:={x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{3}} +\mathrm{37}{x}^{\mathrm{2}} −\mathrm{60}{x}+\mathrm{32}+\mathrm{8}{x}^{\mathrm{2}} −\mathrm{40}{x}+\mathrm{68} \\ $$$$\:\:\:\:=\mathrm{8}{x}^{\mathrm{2}} −\mathrm{40}{x}+\mathrm{68} \\ $$$$\:\:\:\:=\mathrm{8}\left({x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{10}\right)−\mathrm{12} \\ $$$$\:\:\:\:=\mathrm{8}{t}−\mathrm{12} \\ $$$${t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{12}=\mathrm{0} \\ $$$$\left({t}−\mathrm{2}\right)\left({t}−\mathrm{6}\right)=\mathrm{0} \\ $$$${t}_{\mathrm{1},\mathrm{2}} =\mathrm{2},\:\mathrm{6}\:\Rightarrow\frac{\mathrm{1}}{{t}_{\mathrm{1},\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}_{\mathrm{1},\mathrm{2}} ^{\mathrm{2}} −\mathrm{5}{x}_{\mathrm{1},\mathrm{2}} +\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}_{\mathrm{3},\mathrm{4}} ^{\mathrm{2}} −\mathrm{5}{x}_{\mathrm{3},\mathrm{4}} +\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${i}.{e}. \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{5}{a}+\mathrm{10}}+\frac{\mathrm{1}}{{b}^{\mathrm{2}} −\mathrm{5}{b}+\mathrm{10}}+\frac{\mathrm{1}}{{c}^{\mathrm{2}} −\mathrm{5}{c}+\mathrm{10}}+\frac{\mathrm{1}}{{d}^{\mathrm{2}} −\mathrm{5}{d}+\mathrm{10}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\:\checkmark \\ $$

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