Quadratic EquationHard

Question

Let $a,b$ are two different positive integers and the two quadratic equations $(a - 1)x^{2} - \left( a^{2} + 2 \right)x + \left( a^{2} + 2a \right) = 0$ and $(b - 1)x^{2} - \left( b^{2} + 2 \right)x + \left( b^{2} + 2b \right) = 0$ have one common root. Then the value of $\begin{matrix} a^{b} + b^{a} \\ a^{- b} + b^{- a} \end{matrix}$ is equal to

Options

A.256
B.64
C.16
D.72

Solution

$(a - 1)x^{2} - \left( a^{2} - a \right)x - (a + 2)x + a^{2} + 2a = 0$

$$\lbrack(a - 1)x - (a + 2)\rbrack(x - a) = 0$$

The roots of equations are $\frac{a + 2}{a - 1},a$ and $\frac{b + 2}{b - 1},b$

For common root

$$\begin{matrix} \frac{a + 2}{a - 1} = b & \Rightarrow & ab - b - a \\ & \Rightarrow & (a - 1)(b - 1) = 3 \\ & \Rightarrow & (a,b) = (2,4)\text{~or~}(4,2) \\ \therefore & \frac{a^{b} + b^{a}}{\frac{1}{a^{b}} + \frac{1}{b^{a}}} & \\ & & = a^{b}b^{a} = 2^{4}4^{2} = 256 \end{matrix}$$

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