a-3-x-3-x-2-a-1-x-7-0-is-a-cubic-polynomial-in-x-whose-Roots-are-positive-real-numbers-satisfying-225-2-2-7-144-2-2-7-100-2-2-7-Find-a-1-

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Question Number 194991 by York12 last updated on 21/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 Frix last updated on 21/Jul/23

α:β:γ=21:35:189 α=(7/3)∧β=((35)/9)∧γ=21 a_3 =(9/(245))∧a_1 =((77)/(15))

$$\alpha:\beta:\gamma=\mathrm{21}:\mathrm{35}:\mathrm{189} \\ $$$$\alpha=\frac{\mathrm{7}}{\mathrm{3}}\wedge\beta=\frac{\mathrm{35}}{\mathrm{9}}\wedge\gamma=\mathrm{21} \\ $$$${a}_{\mathrm{3}} =\frac{\mathrm{9}}{\mathrm{245}}\wedge{a}_{\mathrm{1}} =\frac{\mathrm{77}}{\mathrm{15}} \\ $$

Commented by York12 last updated on 21/Jul/23