Quantity B

\(\sqrt{9} = 3\)


Quantity A

\(\sqrt{3} + \sqrt{6}\)

can be rewritten as

\(\sqrt{3} + \sqrt{3}\sqrt{2}\)

Now factorizing it by removing the common factor \(\sqrt{3}\) outside we get

\(\sqrt{3}(1+\sqrt{2})\)

Now at this stage, I will write the approximate values for \(\sqrt{3}\) and \(\sqrt{2}\). You should know the square roots at least up to 10.

\(1.732(1+1.414)\)

\(1.732(2.414)\)

Now, I can approximate \(1.732\) as \(1.5\) and \(2.414\) as \(2.0\), thus reducing their original values and get

\(1.5(2.0) = 3\)

But the original numbers are \(1.732\) and \(2.414\) and therefore the result of the product of these two will be greater than \(3\), which is the value of Quantity B

Thus Quantity A is greater than Quantity B

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