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Thursday, January 17, 2019

Positive real numbers and their equations

Given : \displaystyle a,b are positive real numbers such that \displaystyle a\sqrt{a}+b\sqrt{b}=183 and \displaystyle a\sqrt{b}+b\sqrt{a}=182 .
Find : \displaystyle \frac{9}{5}\ (\ a+b\ )

Solution

\displaystyle a\sqrt{a}+b\sqrt{b}=183

\displaystyle {{(\ \sqrt{a}\ )}^{3}}+{{(\ \sqrt{b}\ )}^{3}}=183

\displaystyle {{(\ \sqrt{a}+\sqrt{b}\ )}^{3}}-3\sqrt{{ab}}(\ \sqrt{a}+\sqrt{b}\ )=183\ ........(1)

\displaystyle a\sqrt{b}+b\sqrt{a}=182 

\displaystyle \sqrt{{ab}}\ (\ \sqrt{a}+\sqrt{b}\ )=182\ \ .............(2) 

\displaystyle \text{From}\ (1)\ \text{and}\ (2),

\displaystyle {{(\ \sqrt{a}+\sqrt{b}\ )}^{3}}-546=183

\displaystyle \ \ \ \ \ \ \ \ {{(\ \sqrt{a}+\sqrt{b}\ )}^{3}}=729 

\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{a}+\sqrt{b}=9\ \ \ .............(3)

\displaystyle \text{From}\ (2)\ \text{and}\ (3),

\displaystyle \ \ \ \ \ \ \ \ \ \ \ \sqrt{{ab}}=\frac{{182}}{9}

\displaystyle {{(\ \sqrt{a}+\sqrt{b}\ )}^{2}}=81\ \ \ \ \left[ {By\ (3)} \right]

\displaystyle a+b+2\sqrt{{ab}}=81

\displaystyle \ \ a+b+\frac{{364}}{9}=81

\displaystyle \ a+b=\frac{{365}}{9}=73\times \frac{5}{9}

\displaystyle \therefore \ \ \frac{9}{5}\ (\ a+b\ )=73

 

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