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Question Number 195027 by York12 last updated on 22/Jul/23

a_3 x^3 −x^2 +a_1 x−7=0 is a cubic polynomial in x whose Roots are α , β , γ positive real numbers satisfying ((225α^2 )/(α^2 +7))=((144β^2 )/(β^2 +7))=((100γ^2 )/(γ^2 +7)) find (a_1 )

$${a}_{\mathrm{3}} {x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}−\mathrm{7}=\mathrm{0}\:{is}\:{a}\:{cubic}\:{polynomial}\:{in}\:{x} \\ $$$${whose}\:{Roots}\:{are}\:\alpha\:,\:\beta\:,\:\gamma\:{positive}\:{real}\:{numbers} \\ $$$${satisfying} \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma^{\mathrm{2}} +\mathrm{7}} \\ $$$${find}\:\left({a}_{\mathrm{1}} \right) \\ $$

Commented by mr W last updated on 25/Jul/23

i got a_1 =((77)/(15))

$${i}\:{got}\:{a}_{\mathrm{1}} =\frac{\mathrm{77}}{\mathrm{15}} \\ $$

Commented by York12 last updated on 25/Jul/23

please sir explain why

$${please}\:{sir}\:{explain}\:{why} \\ $$

Answered by mr W last updated on 25/Jul/23

⇒α+β+γ=(1/a_3 ) ((1/x))^3 −(a_1 /7)((1/x))^2 +(1/7)((1/x))−(a_3 /7)=0 ⇒(a_1 /7)=(1/α)+(1/β)+(1/γ) a_3 x^3 −x^2 +a_1 x−7=0 x^2 +7=x(a_3 x^2 +a_1 ) ((225α^2 )/(α^2 +7))=((144β^2 )/(β^2 +7))=((100γ^2 )/(γ^2 +7))=(1/k), say ((225α^2 )/(α(a_3 α^2 +a_1 )))=((144β^2 )/(β(a_3 β^2 +a_1 )))=((100γ^2 )/(γ(a_3 γ^2 +a_1 )))=(1/k) ⇒a_3 α+(a_1 /α)=225k ⇒a_3 β+(a_1 /β)=144k ⇒a_3 γ+(a_1 /γ)=100k ⇒a_3 (α+β+γ)+a_1 ((1/α)+(1/β)+(1/γ))=469k ⇒1+(a_1 ^2 /7)=469k ⇒a_1 =(√(7(469k−1))) ((225)/(1+(7/α^2 )))=((144)/(1+(7/β^2 )))=((100)/(1+(7/γ^2 )))=(1/k) ⇒((√7)/α)=(√(225k−1)) ⇒((√7)/β)=(√(144k−1)) ⇒((√7)/γ)=(√(100k−1)) (√7)((1/α)+(1/β)+(1/γ))=(√(225k−1))+(√(144k−1))+(√(100k−1)) (a_1 /( (√7)))=(√(225k−1))+(√(144k−1))+(√(100k−1)) ⇒(√(469k−1))=(√(225k−1))+(√(144k−1))+(√(100k−1)) ⇒(√(225k−1))+(√(144k−1))=(√(469k−1))−(√(100k−1)) ⇒(√((225k−1)(144k−1)))=100k−(√((469k−1)(100k−1))) ⇒245k−2=2(√((469k−1)(100k−1))) ⇒127575k=1296 ⇒k=((1296)/(127575))=((16)/(1575)) a_1 =(√(7(((469×16)/(1575))−1)))=((77)/(15)) ✓ α=(√(7/(225×((16)/(1575))−1)))=(7/3) β=(√(7/(144×((16)/(1575))−1)))=((35)/9) γ=(√(7/(100×((16)/(1575))−1)))=21 a_3 =(1/((7/3)+((35)/9)+21))=(9/(245))

$$\Rightarrow\alpha+\beta+\gamma=\frac{\mathrm{1}}{{a}_{\mathrm{3}} } \\ $$$$\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\frac{{a}_{\mathrm{1}} }{\mathrm{7}}\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{7}}\left(\frac{\mathrm{1}}{{x}}\right)−\frac{{a}_{\mathrm{3}} }{\mathrm{7}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{a}_{\mathrm{1}} }{\mathrm{7}}=\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\gamma} \\ $$$$ \\ $$$${a}_{\mathrm{3}} {x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}−\mathrm{7}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\mathrm{7}={x}\left({a}_{\mathrm{3}} {x}^{\mathrm{2}} +{a}_{\mathrm{1}} \right) \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{1}}{{k}},\:{say} \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha\left({a}_{\mathrm{3}} \alpha^{\mathrm{2}} +{a}_{\mathrm{1}} \right)}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta\left({a}_{\mathrm{3}} \beta^{\mathrm{2}} +{a}_{\mathrm{1}} \right)}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma\left({a}_{\mathrm{3}} \gamma^{\mathrm{2}} +{a}_{\mathrm{1}} \right)}=\frac{\mathrm{1}}{{k}} \\ $$$$\Rightarrow{a}_{\mathrm{3}} \alpha+\frac{{a}_{\mathrm{1}} }{\alpha}=\mathrm{225}{k} \\ $$$$\Rightarrow{a}_{\mathrm{3}} \beta+\frac{{a}_{\mathrm{1}} }{\beta}=\mathrm{144}{k} \\ $$$$\Rightarrow{a}_{\mathrm{3}} \gamma+\frac{{a}_{\mathrm{1}} }{\gamma}=\mathrm{100}{k} \\ $$$$\Rightarrow{a}_{\mathrm{3}} \left(\alpha+\beta+\gamma\right)+{a}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\gamma}\right)=\mathrm{469}{k} \\ $$$$\Rightarrow\mathrm{1}+\frac{{a}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{7}}=\mathrm{469}{k} \\ $$$$\Rightarrow{a}_{\mathrm{1}} =\sqrt{\mathrm{7}\left(\mathrm{469}{k}−\mathrm{1}\right)} \\ $$$$ \\ $$$$\frac{\mathrm{225}}{\mathrm{1}+\frac{\mathrm{7}}{\alpha^{\mathrm{2}} }}=\frac{\mathrm{144}}{\mathrm{1}+\frac{\mathrm{7}}{\beta^{\mathrm{2}} }}=\frac{\mathrm{100}}{\mathrm{1}+\frac{\mathrm{7}}{\gamma^{\mathrm{2}} }}=\frac{\mathrm{1}}{{k}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{7}}}{\alpha}=\sqrt{\mathrm{225}{k}−\mathrm{1}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{7}}}{\beta}=\sqrt{\mathrm{144}{k}−\mathrm{1}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{7}}}{\gamma}=\sqrt{\mathrm{100}{k}−\mathrm{1}} \\ $$$$\sqrt{\mathrm{7}}\left(\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\gamma}\right)=\sqrt{\mathrm{225}{k}−\mathrm{1}}+\sqrt{\mathrm{144}{k}−\mathrm{1}}+\sqrt{\mathrm{100}{k}−\mathrm{1}} \\ $$$$\frac{{a}_{\mathrm{1}} }{\:\sqrt{\mathrm{7}}}=\sqrt{\mathrm{225}{k}−\mathrm{1}}+\sqrt{\mathrm{144}{k}−\mathrm{1}}+\sqrt{\mathrm{100}{k}−\mathrm{1}} \\ $$$$\Rightarrow\sqrt{\mathrm{469}{k}−\mathrm{1}}=\sqrt{\mathrm{225}{k}−\mathrm{1}}+\sqrt{\mathrm{144}{k}−\mathrm{1}}+\sqrt{\mathrm{100}{k}−\mathrm{1}} \\ $$$$\Rightarrow\sqrt{\mathrm{225}{k}−\mathrm{1}}+\sqrt{\mathrm{144}{k}−\mathrm{1}}=\sqrt{\mathrm{469}{k}−\mathrm{1}}−\sqrt{\mathrm{100}{k}−\mathrm{1}} \\ $$$$\Rightarrow\sqrt{\left(\mathrm{225}{k}−\mathrm{1}\right)\left(\mathrm{144}{k}−\mathrm{1}\right)}=\mathrm{100}{k}−\sqrt{\left(\mathrm{469}{k}−\mathrm{1}\right)\left(\mathrm{100}{k}−\mathrm{1}\right)} \\ $$$$\Rightarrow\mathrm{245}{k}−\mathrm{2}=\mathrm{2}\sqrt{\left(\mathrm{469}{k}−\mathrm{1}\right)\left(\mathrm{100}{k}−\mathrm{1}\right)} \\ $$$$\Rightarrow\mathrm{127575}{k}=\mathrm{1296} \\ $$$$\Rightarrow{k}=\frac{\mathrm{1296}}{\mathrm{127575}}=\frac{\mathrm{16}}{\mathrm{1575}} \\ $$$${a}_{\mathrm{1}} =\sqrt{\mathrm{7}\left(\frac{\mathrm{469}×\mathrm{16}}{\mathrm{1575}}−\mathrm{1}\right)}=\frac{\mathrm{77}}{\mathrm{15}}\:\checkmark \\ $$$$ \\ $$$$\alpha=\sqrt{\frac{\mathrm{7}}{\mathrm{225}×\frac{\mathrm{16}}{\mathrm{1575}}−\mathrm{1}}}=\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$\beta=\sqrt{\frac{\mathrm{7}}{\mathrm{144}×\frac{\mathrm{16}}{\mathrm{1575}}−\mathrm{1}}}=\frac{\mathrm{35}}{\mathrm{9}} \\ $$$$\gamma=\sqrt{\frac{\mathrm{7}}{\mathrm{100}×\frac{\mathrm{16}}{\mathrm{1575}}−\mathrm{1}}}=\mathrm{21} \\ $$$${a}_{\mathrm{3}} =\frac{\mathrm{1}}{\frac{\mathrm{7}}{\mathrm{3}}+\frac{\mathrm{35}}{\mathrm{9}}+\mathrm{21}}=\frac{\mathrm{9}}{\mathrm{245}} \\ $$

Commented by York12 last updated on 25/Jul/23

thanks so much

$${thanks}\:{so}\:{much} \\ $$

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