MHB Solving for Common Root in $(1)$ and $(2)$

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The discussion focuses on finding a common root between two quadratic equations, where the coefficients depend on distinct natural numbers a and b, both greater than 1. The equations are structured to reveal a relationship between the coefficients that leads to a shared root. Participants explore the implications of this shared root on the values of a and b, ultimately aiming to compute the expression a^a + b^b divided by a^(-b) + b^(-a). The problem emphasizes the conditions that a and b must satisfy, particularly that they are not equal and both exceed 1. The solution involves algebraic manipulation and understanding of quadratic roots.
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$(a-1)x^2-(a^2+2)x+(a^2+2a)=0----(1)\\
(b-1)x^2-(b^2+2)x+(b^2+2b)=0----(2)
$
if $(1)$ and $(2)$ have one root in common ,
(here $a,b\in N$ ,$a\neq b,\,\, and \,\, a>1,b>1$)
find value of :
$\dfrac{a^a+b^b}{a^{-b}+b^{-a}}$
 
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Albert said:
$(a-1)x^2-(a^2+2)x+(a^2+2a)=0----(1)\\
(b-1)x^2-(b^2+2)x+(b^2+2b)=0----(2)
$
if $(1)$ and $(2)$ have one root in common ,
(here $a,b\in N$ ,$a\neq b,\,\, and \,\, a>1,b>1$)
find value of :
$\dfrac{a^a+b^b}{a^{-b}+b^{-a}}$
hint
(1) and (2) can be factorized
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

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