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Olympiad workbook
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Olympiad workbook
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IOQM
Chapter
Quadratic Equation

Olympiad workbook · IOQM · Quadratic Equation

Suppose a, b are positive real numbers such that a a +b b =183, a b...

Suppose a,ba, b are positive real numbers such that aa+bb=183,ab+ba=182a \sqrt{a}+b \sqrt{b}=183, a \sqrt{b}+b \sqrt{a}=182. Find 95(a+b)\frac{9}{5}(a+b).

Correct answer

73

Step-by-step solution

aa+bb=183a \sqrt{a}+b \sqrt{b}=183 and ab+ba=182a \sqrt{b}+b \sqrt{a}=182 we have to find 95(a+b)\frac{9}{5}(a+b) Let a=A2,b=B2a=A^2, b=B^2 A3+B3=183…………. (i) A2B+B2A=182……….. (ii) (1)+(2)A3+B3+3AB(A+B)=183+3×182(A+B)3=183+546(A+B)3=729⇒A+B=9\begin{aligned} & A^3+B^3=183 \ldots \ldots \ldots \ldots . \text { (i) } \\ & A^2 B+B^2 A=182 \ldots \ldots \ldots . . \text { (ii) } \\ & (1)+(2) \\ & A^3+B^3+3 A B(A+B)=183+3 \times 182 \\ & (A+B)^3=183+546 \\ & (A+B)^3=729 \\ & \Rightarrow A+B=9 \end{aligned}  from (2)→AB(A+B)=182AB=1829a+b=A2+B2=(A+B)2−2AB=81−3649=365995×3659=73\begin{aligned} & \text { from }(2) \rightarrow A B(A+B)=182 \\ & A B=\frac{182}{9} \\ & a+b=A^2+B^2=(A+B)^2-2 A B \\ & =81-\frac{364}{9}=\frac{365}{9} \\ & \frac{9}{5} \times \frac{365}{9}=73\end{aligned}
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