Kuwait
Let ^@a^@ and ^@b^@ be positive real numbers such that
^@\dfrac{1}{a} - \dfrac{1}{b} - \dfrac{1}{ a + b } = 0^@
Find the value of ^@\left(\dfrac{b}{a} + \dfrac{a}{b} \right)^2^@.
Answer:
^@5^@
- We are given,
^@\dfrac{1}{a} - \dfrac{1}{b} - \dfrac{1}{ a + b } = 0^@
Let ^@x = \dfrac{b}{a}^@
^@\implies b = ax^@ - Now,
^@\begin{align} & \implies \dfrac{1}{a} - \dfrac{1}{b} - \dfrac{1}{ a + b } = 0 \\ & \implies \dfrac{1}{a} - \dfrac{1}{ ax } - \dfrac{1}{ a + ax } = 0 \\ & \implies \dfrac{1}{a}\left(1 - \dfrac{1}{ x } - \dfrac{1}{ 1 + x } \right) = 0 \\ & \implies \dfrac{1}{a}\left( \dfrac{ x(1 + x) - (1 + x) - x }{ x(1 + x) } \right) = 0 \\ & \implies \dfrac{1}{a}\left( \dfrac{ x + x^2 - 1 - x - x }{ x(1 + x) } \right) = 0 \\ & \implies \dfrac{1}{a}\left( \dfrac{ x^2 - x - 1 }{ x(1 + x) } \right) = 0 \\ & \implies \dfrac{ x^2 - x - 1 }{ x(1 + x) } = 0 && (\because a > 0) \\ & \implies x^2 - x - 1 = 0 \\ & \implies x = \dfrac{ 1 \pm \sqrt{ 1 + 4 } }{ 2(1) } \\ & \implies x = \dfrac{ 1 \pm \sqrt{5} }{ 2 } \\ & \text{ Since a and b are positive, x > 0 } \\ & \implies x = \dfrac{ 1 + \sqrt{ 5 } }{ 2 } \end{align}^@ - 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}^@ - Hence, the value of ^@\left(\dfrac{b}{a} + \dfrac{a}{b} \right)^2^@ is ^@5^@.
