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

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Root
AI Thread Summary
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.
Albert1
Messages
1,221
Reaction score
0
$(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 :
$\dfrac{a^a+b^b}{a^{-b}+b^{-a}}$
 
Mathematics news on Phys.org
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
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top