SOLUTION: Let a and b be the roots of the quadratic x^2 - 5x + 3 = 0. Find the quadratic whose roots are a^2/b and b^2/a. x^2 + ___ x + ___

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Let a and b be the roots of the quadratic x^2 - 5x + 3 = 0. Find the quadratic whose roots are a^2/b and b^2/a. x^2 + ___ x + ___      Log On


   

Question 1209227: Let a and b be the roots of the quadratic x^2 - 5x + 3 = 0. Find the quadratic whose roots are a^2/b and b^2/a.
x^2 + ___ x + ___
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let a and b be the roots of the quadratic x^2 - 5x + 3 = 0.
Find the quadratic whose roots are a^2/b and b^2/a.
x^2 + ___ x + ___
~~~~~~~~~~~~~~~~~~~~~~~~~ 

According to Vieta's theorem, we have

    a + b = 5,

    ab = 3.


Next,  a%5E2%2Fb + b%5E2%2Fa = %28a%5E3%2Bb%5E3%29%2Fab.


In the numerator,  

    a^3 + b^3 = (a+b)*(a^2-ab+b^2) = (a+b)*((a^2 +2ab+b^2)-3ab) = (a+b)*((a+b^2-3ab) = (a+b)^3 - 3ab*(a+b) =

                substitute here a+b = 5, ab = 3 and continue

              = 5^3 - 3*3*5 = 125-45 = 80.


Therefore,  

    a%5E2%2Fb + b%5E2%2Fa = %28a%5E3%2Bb%5E3%29%2Fab = 80%2F3.


Also,  a%5E2%2Fb.b%5E2%2Fa = %28a%5E2%2Ab%5E2%29%2F%28ab%29 = ab = 3.


So, due to Vieta's theorem (again), the desired quadratic equation has 
the coefficient  -80%2F3  at x  and  the constant term 3.


ANSWER.  The desired quadratic equation is  x^2 - %2880%2F3%29x + 3 = 0.

Solved.