If \((a+b)^2 = 5 + 2\sqrt{6}\), what can be the possible value of ‘\(b\)’ from the following?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\sqrt{3}\)
An expression of the form \(p+2\sqrt{q}\) is a perfect square \((\sqrt{m}+\sqrt{n})^2=m+n+2\sqrt{mn}\) when two numbers have sum \(p\) and product \(q\). Here the sum must be 5 and the product 6.
Numbers with sum 5 and product 6: 2 and 3
\((\sqrt{2}+\sqrt{3})^2=2+3+2\sqrt{6}=5+2\sqrt{6}\)
So \(a\) and \(b\) are \(\sqrt{2}\) and \(\sqrt{3}\), in either order
Of these two values, only \(\sqrt{3}\) is among the options.