Question Number 177648 by BaliramKumar last updated on 07/Oct/22
![If ((a + b)/( (√(ab)))) = (4/1) then a : b = ? [ a>b ]](https://www.tinkutara.com/question/Q177648.png)
$${If}\:\:\frac{{a}\:+\:{b}}{\:\sqrt{{ab}}}\:=\:\frac{\mathrm{4}}{\mathrm{1}}\:{then}\:\:\:\:{a}\::\:{b}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\:{a}>{b}\:\right] \\ $$
Answered by Rasheed.Sindhi last updated on 07/Oct/22
![a+b=4(√(ab)) a^2 +b^2 =16ab−2ab=14ab ((a )/b)+(b/a)=14 x+(1/x)=14; [ (a/b)=x] x^2 −14x+1=0 x=((14±(√(196−4)))/2)=((14±8(√3))/2)=7±4(√3) (a/b)=((7±4(√3))/1) ∵ a>b ∴ a:b=(7+4(√3) ):1](https://www.tinkutara.com/question/Q177654.png)
$${a}+{b}=\mathrm{4}\sqrt{{ab}}\: \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{16}{ab}−\mathrm{2}{ab}=\mathrm{14}{ab} \\ $$$$\frac{{a}\:}{{b}}+\frac{{b}}{{a}}=\mathrm{14} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{14};\:\left[\:\frac{{a}}{{b}}={x}\right] \\ $$$${x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{14}\pm\sqrt{\mathrm{196}−\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{14}\pm\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{7}\pm\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\frac{{a}}{{b}}=\frac{\mathrm{7}\pm\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{1}} \\ $$$$\because\:{a}>{b} \\ $$$$\therefore\:\:{a}:{b}=\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\:\right):\mathrm{1} \\ $$
Commented by BaliramKumar last updated on 07/Oct/22
$${nice}\:{solution} \\ $$