Let a and b be positive real numbers such that
1a−1b−1a+b=0
Find the value of (ba+ab)2.


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Answer:

5

Step by Step Explanation:
  1. We are given,
    1a−1b−1a+b=0
    Let x=ba
    ⟹b=ax
  2. Now,
    ⟹1a−1b−1a+b=0⟹1a−1ax−1a+ax=0⟹1a(1−1x−11+x)=0⟹1a(x(1+x)−(1+x)−xx(1+x))=0⟹1a(x+x2−1−x−xx(1+x))=0⟹1a(x2−x−1x(1+x))=0⟹x2−x−1x(1+x)=0(∵
  3. Now,
    \begin{align} \left( \dfrac{b}{a} + \dfrac{a}{b} \right)& = \left(x + \dfrac{1}{x} \right)^2 \\ & = \left(\dfrac{ 1 + \sqrt{ 5 } }{ 2 } + \dfrac{ 2 }{ 1 + \sqrt{ 5 } } \right)^2 \\ & = \left(\dfrac{ 1 + \sqrt{ 5 } }{ 2 } + \dfrac{ 2 }{ 1 + \sqrt{ 5 } } \times \dfrac{ 1 - \sqrt{ 5 } }{ 1 - \sqrt{ 5 } } \right)^2 \\ & = \left(\dfrac{ 1 + \sqrt{ 5 } }{ 2 } + \dfrac{ 2 - 2\sqrt{5} }{ -4 } \right)^2 \\ & = \left(\dfrac{ 1 + \sqrt{ 5 } }{ 2 } - \dfrac{1}{2} + \dfrac{ \sqrt{5} }{ 2 } \right)^2 \\ & = \left( \sqrt{5} \right)^2 \\ & = 5 \end{align}
  4. Hence, the value of \left(\dfrac{b}{a} + \dfrac{a}{b} \right)^2 is 5.

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