Question

Suppose a,b are positive real numbers such that a√a+b√b=183. a√b+b√a=182. Find 9÷5(a+b)

Answer 1

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Answer 2

√a +b√b=183,

a√b +b√a=182 or √(ab)(√a+√b)=182.  

Let’s assume p=√a  

and q=√b.  

We got p³+q³=183 ——-(a)  

& pq(p+q)=182 Or  

3pq(p+q)=

3 x 182=546 ——-(b).  

Add (a) & (b) to obtain,

$\begin{array}{l}{p^{3}+q^{3}+3 p q(p+q)=(p+q)^{3}} \\ {=183+546=729=9^{3}}\end{array}$

We then claim that, p+q=9.

Going back to equation (a),  

We have:  

$\begin{array}{l}{p^{3}+q^{3}=183 \text { or }(p+q)\left(p^{2}+q^{2}-p q\right)=183 \text { or }} \\ {(p+q)\left(p^{2}+q^{2}\right)-p q(p+q)=183 \text { or }} \\ {(p+q)\left(p^{2}+q^{2}\right)-182=183 \text { or }} \\ {(p+q)\left(p^{2}+q^{2}\right)=365 \text { or }}\end{array}$

$\left(\mathrm{p}^{2}+\mathrm{q}^{2}\right)=365 /(\mathrm{p}+\mathrm{q})=365 / 9$

Then the required expression = (9/5) (a + b) = $(9 / 5)\left(\mathrm{p}^{2}+\mathrm{q}^{2}\right)$

=(9/5)(365/9) = 73.