a-b-ab-4-then-a-b-

Question Number 179660 by mathlove last updated on 31/Oct/22

$$\frac{{a}+{b}}{\:\sqrt{{ab}}}=\mathrm{4}\:\:\:\:{then}\:{a}:{b}=? \\ $$

Answered by FelipeLz last updated on 31/Oct/22

((a+b)/( (√(ab)))) = 4 ((a^2 +2ab+b^2 )/( ab)) = 16 (a/b)+2+(b/a) = 16 (a/b)+(b/a) = 14 (a/b) = x → x+(1/x) = 14 x^2 −14x+1 = 0 → x = ((14±8(√3))/2) = 7±4(√3)

$$\frac{{a}+{b}}{\:\sqrt{{ab}}}\:=\:\mathrm{4} \\ $$$$\frac{{a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} }{\:{ab}}\:=\:\mathrm{16} \\ $$$$\frac{{a}}{{b}}+\mathrm{2}+\frac{{b}}{{a}}\:=\:\mathrm{16} \\ $$$$\frac{{a}}{{b}}+\frac{{b}}{{a}}\:=\:\mathrm{14} \\ $$$$\frac{{a}}{{b}}\:=\:{x}\:\rightarrow\:{x}+\frac{\mathrm{1}}{{x}}\:=\:\mathrm{14}\: \\ $$$${x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{1}\:=\:\mathrm{0}\:\rightarrow\:{x}\:=\:\frac{\mathrm{14}\pm\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:\mathrm{7}\pm\mathrm{4}\sqrt{\mathrm{3}} \\ $$

Answered by Rasheed.Sindhi last updated on 31/Oct/22

((a+b)/( (√(ab))))=4 then a:b=? ((((√a) )^2 )/( (√a) (√b) ))+((((√b) )^2 )/( (√a) (√b) ))=4 (((√a) )/( (√b) ))+((√b)/( (√a) ))=4 x+(1/x)=4 [Let (((√a) )/( (√b) ))=x] x^2 −4x+1=0 x=((4±(√(16−4)) )/2)=2±(√3) ((√a)/( (√b) ))=2±(√3) (a/( b ))=(2±(√3) )^2 =4+3±4(√3) =7±4(√3) a:b=7±4(√3) :1

$$\frac{{a}+{b}}{\:\sqrt{{ab}}}=\mathrm{4}\:\:\:\:{then}\:{a}:{b}=? \\ $$$$\frac{\left(\sqrt{{a}}\:\right)^{\mathrm{2}} \:}{\:\sqrt{{a}}\:\sqrt{{b}}\:}+\frac{\left(\sqrt{{b}}\:\right)^{\mathrm{2}} }{\:\sqrt{{a}}\:\sqrt{{b}}\:}=\mathrm{4} \\ $$$$\frac{\sqrt{{a}}\:}{\:\sqrt{{b}}\:}+\frac{\sqrt{{b}}}{\:\sqrt{{a}}\:}=\mathrm{4} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{4}\:\:\left[{Let}\:\frac{\sqrt{{a}}\:}{\:\sqrt{{b}}\:}={x}\right] \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{4}}\:}{\mathrm{2}}=\mathrm{2}\pm\sqrt{\mathrm{3}}\: \\ $$$$\frac{\sqrt{{a}}}{\:\sqrt{{b}}\:}=\mathrm{2}\pm\sqrt{\mathrm{3}}\: \\ $$$$\frac{{a}}{\:{b}\:}=\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} =\mathrm{4}+\mathrm{3}\pm\mathrm{4}\sqrt{\mathrm{3}}\:=\mathrm{7}\pm\mathrm{4}\sqrt{\mathrm{3}} \\ $$$${a}:{b}=\mathrm{7}\pm\mathrm{4}\sqrt{\mathrm{3}}\::\mathrm{1} \\ $$

Commented by mathlove last updated on 01/Nov/22

$${thanks} \\ $$

Answered by Rasheed.Sindhi last updated on 01/Nov/22

determinant (((((a+b)/( (√(ab))))=4 then a:b=?))) a=4(√(ab)) −b a=(√b) (4(√a) −(√b) ) (a/( (√b)))=4(√a) −(√b) (a/b)=((4(√a) −(√b) )/( (√b) ))=4((√(a/b)) )−1 ((√(a/b)) )^2 −4((√(a/b)) )+1=0 (√(a/b)) =((4±(√(16−4)) )/2)=2±(√3) (a/b)=(2±(√3) )^2 =4+3±4(√3) =7±4(√3)

$$\begin{array}{|c|}{\frac{\boldsymbol{{a}}+\boldsymbol{{b}}}{\:\sqrt{\boldsymbol{{ab}}}}=\mathrm{4}\:\:\:\:\boldsymbol{{then}}\:\boldsymbol{{a}}:\boldsymbol{{b}}=?}\\\hline\end{array} \\ $$$${a}=\mathrm{4}\sqrt{{ab}}\:−{b} \\ $$$${a}=\sqrt{{b}}\:\left(\mathrm{4}\sqrt{{a}}\:−\sqrt{{b}}\:\right) \\ $$$$\frac{{a}}{\:\sqrt{{b}}}=\mathrm{4}\sqrt{{a}}\:−\sqrt{{b}}\: \\ $$$$\frac{{a}}{{b}}=\frac{\mathrm{4}\sqrt{{a}}\:−\sqrt{{b}}\:}{\:\sqrt{{b}}\:}=\mathrm{4}\left(\sqrt{\frac{{a}}{{b}}}\:\right)−\mathrm{1} \\ $$$$\left(\sqrt{\frac{{a}}{{b}}}\:\right)^{\mathrm{2}} −\mathrm{4}\left(\sqrt{\frac{{a}}{{b}}}\:\right)+\mathrm{1}=\mathrm{0} \\ $$$$\sqrt{\frac{{a}}{{b}}}\:=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{4}}\:}{\mathrm{2}}=\mathrm{2}\pm\sqrt{\mathrm{3}}\: \\ $$$$\frac{{a}}{{b}}=\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} =\mathrm{4}+\mathrm{3}\pm\mathrm{4}\sqrt{\mathrm{3}}\:=\mathrm{7}\pm\mathrm{4}\sqrt{\mathrm{3}}\: \\ $$

Commented by mathlove last updated on 04/Nov/22

$${thanks} \\ $$

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