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Question-207984




Question Number 207984 by Thomaseinstein last updated on 02/Jun/24
Answered by Frix last updated on 02/Jun/24
p, q >0  (1/p^2 )−(1/q^2 )=(4/3) ⇒ q^2 =((3p^2 )/( 3−4p^2 ))  p+q=6 ⇒ q^2 =(p−6)^2   ((3p^2 )/( 3−4p^2 ))=(p−6)^2   p^4 −12p^3 +36p^2 +9p−27=0  p≈.854016106031  ⇒  q≈5.14598389396  ⇒  m≈.729343509360  n≈26.4811502370
$${p},\:{q}\:>\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{p}^{\mathrm{2}} }−\frac{\mathrm{1}}{{q}^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\:{q}^{\mathrm{2}} =\frac{\mathrm{3}{p}^{\mathrm{2}} }{\:\mathrm{3}−\mathrm{4}{p}^{\mathrm{2}} } \\ $$$${p}+{q}=\mathrm{6}\:\Rightarrow\:{q}^{\mathrm{2}} =\left({p}−\mathrm{6}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{3}{p}^{\mathrm{2}} }{\:\mathrm{3}−\mathrm{4}{p}^{\mathrm{2}} }=\left({p}−\mathrm{6}\right)^{\mathrm{2}} \\ $$$${p}^{\mathrm{4}} −\mathrm{12}{p}^{\mathrm{3}} +\mathrm{36}{p}^{\mathrm{2}} +\mathrm{9}{p}−\mathrm{27}=\mathrm{0} \\ $$$${p}\approx.\mathrm{854016106031} \\ $$$$\Rightarrow \\ $$$${q}\approx\mathrm{5}.\mathrm{14598389396} \\ $$$$\Rightarrow \\ $$$${m}\approx.\mathrm{729343509360} \\ $$$${n}\approx\mathrm{26}.\mathrm{4811502370} \\ $$

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