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last updated by  Ashani Dasgupta 5 months ago
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• #21685
bibhu
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solve for a & b?

#21692
Ashani Dasgupta
Keymaster

Is the question complete?

For example, if we plug in a = 1, then we will get (possibly complex) solutions of b.

Computation:

if a = 1

$$(1-b)(1 – \sqrt{b} ) = 1$$

Expanding we get $$1 – \sqrt{b} – b + b \sqrt{b} = 1$$

This implies $$b \sqrt b = b + \sqrt b \Rightarrow b = \sqrt b + 1$$

Set $$\sqrt b =x$$ and solve the qudratic.

This process will work for any value of a.

This problem reminds me of a problem from Pre RMO 2017. If you expand the given expression you will get $$a \sqrt a + b \sqrt b – a \sqrt b – b \sqrt a = 1$$. The Pre RMO problem set the values of $$a \sqrt a + b \sqrt b = 183$$ and $$a \sqrt b -+ b \sqrt a = 182$$

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