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Question Number 89991 by jagoll last updated on 20/Apr/20

if a,b > 0 and (a^2 /b^2 ) = (5/3) find ((a^2 +b^2 )/(ab))

$$\mathrm{if}\:\mathrm{a},\mathrm{b}\:>\:\mathrm{0}\:\mathrm{and}\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:=\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$ $$\mathrm{find}\:\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{ab}} \\ $$

Commented byjohn santu last updated on 20/Apr/20

ab = (√(a^2 b^2 )) , since a,b > 0 a^2 = (5/3)b^2 ⇒ab = (√((5b^4 )/3)) = ((b^2 (√(15)))/3) then ((a^2 +b^2 )/(ab)) = (((8b^2 )/3)/((b^2 (√(15)))/3)) = (8/(√(15)))

$${ab}\:=\:\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\:,\:{since}\:{a},{b}\:>\:\mathrm{0} \\ $$ $${a}^{\mathrm{2}} =\:\frac{\mathrm{5}}{\mathrm{3}}{b}^{\mathrm{2}} \:\Rightarrow{ab}\:=\:\sqrt{\frac{\mathrm{5}{b}^{\mathrm{4}} }{\mathrm{3}}}\:=\:\frac{{b}^{\mathrm{2}} \sqrt{\mathrm{15}}}{\mathrm{3}} \\ $$ $${then}\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}}\:=\:\frac{\frac{\mathrm{8}{b}^{\mathrm{2}} }{\mathrm{3}}}{\frac{{b}^{\mathrm{2}} \sqrt{\mathrm{15}}}{\mathrm{3}}} \\ $$ $$=\:\frac{\mathrm{8}}{\sqrt{\mathrm{15}}} \\ $$

Commented bymr W last updated on 20/Apr/20

((a/b))^2 =(5/3) (a/b)=((√5)/(√3)) ((a^2 +b^2 )/(ab))=(a/b)+(b/a)=((√5)/(√3))+((√3)/(√5))=(8/(√(15)))=((8(√(15)))/(15))

$$\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{3}} \\ $$ $$\frac{{a}}{{b}}=\frac{\sqrt{\mathrm{5}}}{\sqrt{\mathrm{3}}} \\ $$ $$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}}=\frac{{a}}{{b}}+\frac{{b}}{{a}}=\frac{\sqrt{\mathrm{5}}}{\sqrt{\mathrm{3}}}+\frac{\sqrt{\mathrm{3}}}{\sqrt{\mathrm{5}}}=\frac{\mathrm{8}}{\sqrt{\mathrm{15}}}=\frac{\mathrm{8}\sqrt{\mathrm{15}}}{\mathrm{15}} \\ $$

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