Question

a√a+b√b=183 and a√b+b√a=182 then find 9/5(a+b)

Answer 1

a√a+b√b=183a√a+b√b=183,

a√b+b√a=182a√b+b√a=182

√(ab)(√a+√b)=182√(ab)(√a+√b)=182

Let’s assume u=√au=√aandv=√v=√b.b.

The given equations are now free of annoying radicals viz.:

u³+v³=183u³+v³=183 ——-(a) &

uv(u+v)=182uv(u+v)=182Or3uv(u+v)=3∗182=5463uv(u+v)=3∗182=546 ——-(b).

Add (a) & (b) to obtain,

u³+v³+3uv(u+v)=(u+v)³=183+546=729=9³u³+v³+3uv(u+v)=(u+v)³=183+546=729=9³.

We then claim that, u+v=9u+v=9.

Going back to eqn (a), we’ve:

u³+v³=183u³+v³=183

or(u+v)(u²+v²−uv)=183(u+v)(u²+v²−uv)=183

or(u+v)(u²+v²)−uv(u+v)=183(u+v)(u²+v²)−uv(u+v)=183

or(u+v)(u²+v²)−182=183(u+v)(u²+v²)−182=183

or(u+v)(u²+v²)=365(u+v)(u²+v²)=365

or(u²+v²)=365/(u+v)=365/9.(u²+v²)=365/(u+v)=365/9.

Then the required expression = (9/5)(a+b)=(9/5)(u²+v²)=(9/5)(365/9)=73.