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Question-200575




Question Number 200575 by sonukgindia last updated on 20/Nov/23
Answered by AST last updated on 20/Nov/23
((ra)/2)=((bx)/2)⇒ra=bx;x=(√(r^2 −a^2 ));b=r−a  ra=bx⇒ra=(r−a)((√(r^2 −a^2 )))  ⇒2r=a+_− a(√(5+4(√2)))⇒a+2b=+_− a(√(5+4(√2)))  ⇒a(1+^− (√(5+4(√2))))=−2b⇒(a/b)=((−2)/(1+^− (√(5+4(√2)))))  ⇒(a/b)≈0.883203
$$\frac{{ra}}{\mathrm{2}}=\frac{{bx}}{\mathrm{2}}\Rightarrow{ra}={bx};{x}=\sqrt{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} };{b}={r}−{a} \\ $$$${ra}={bx}\Rightarrow{ra}=\left({r}−{a}\right)\left(\sqrt{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\mathrm{2}{r}={a}\underset{−} {+}{a}\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}}\Rightarrow{a}+\mathrm{2}{b}=\underset{−} {+}{a}\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow{a}\left(\mathrm{1}\overset{−} {+}\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}}\right)=−\mathrm{2}{b}\Rightarrow\frac{{a}}{{b}}=\frac{−\mathrm{2}}{\mathrm{1}\overset{−} {+}\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow\frac{{a}}{{b}}\approx\mathrm{0}.\mathrm{883203} \\ $$
Commented by mr W last updated on 21/Nov/23
how did you get?  ⇒2r=a+_− a(√(5+4(√2)))
$${how}\:{did}\:{you}\:{get}? \\ $$$$\Rightarrow\mathrm{2}{r}={a}\underset{−} {+}{a}\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}} \\ $$
Commented by AST last updated on 21/Nov/23
((r^2 a^2 )/((r−a)^2 ))=r^2 −a^2 ⇒r^2 a^2 =(r^2 −a^2 )(r^2 +a^2 −2ar)  ⇒r^4 −2ar^3 −a^2 r^2 +2a^3 r−a^4 =0  So,we can solve the polynomial in r, I used   wolfram. r=(1/2)(a+_− a(√(5+4(√2))))
$$\frac{{r}^{\mathrm{2}} {a}^{\mathrm{2}} }{\left({r}−{a}\right)^{\mathrm{2}} }={r}^{\mathrm{2}} −{a}^{\mathrm{2}} \Rightarrow{r}^{\mathrm{2}} {a}^{\mathrm{2}} =\left({r}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\left({r}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{ar}\right) \\ $$$$\Rightarrow{r}^{\mathrm{4}} −\mathrm{2}{ar}^{\mathrm{3}} −{a}^{\mathrm{2}} {r}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{3}} {r}−{a}^{\mathrm{4}} =\mathrm{0} \\ $$$${So},{we}\:{can}\:{solve}\:{the}\:{polynomial}\:{in}\:{r},\:{I}\:{used}\: \\ $$$${wolfram}.\:{r}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}\underset{−} {+}{a}\sqrt{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}}\right) \\ $$
Commented by mr W last updated on 21/Nov/23
thanks sir! i just wanted to know if  you had a way to avoid solving   quadratic equation.
$${thanks}\:{sir}!\:{i}\:{just}\:{wanted}\:{to}\:{know}\:{if} \\ $$$${you}\:{had}\:{a}\:{way}\:{to}\:{avoid}\:{solving}\: \\ $$$${quadratic}\:{equation}. \\ $$

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