Question

What are the roots of the equation (p2-q2)x2 - (q2-r2)x + (r2-p2) = 0

Answer 1

Answer:

The roots are   -1   and    ( p² - r² ) / ( p² - q² )

Step-by-step explanation:

Given the equation

      ( p² - q² ) x² - ( q² - r² ) x + ( r² - p² ) = 0

At a glance, it already looks like x = 1 or x = -1 might be a root.  Indeed, putting x = -1 gives

     ( p² - q² ) (-1)² - ( q² - r² ) (-1) + ( r² - p² )

   = p² - q² + q² - r² + r² - p²

   = 0.

So -1 is a root which means that (x+1) is a factor.  Knowing one factor, it is now easy to write down the factorization:

     ( p² - q² ) x² - ( q² - r² ) x + ( r² - p² ) = 0

=>  ( x + 1 ) ( ( p² - q² ) x  +  ( r² - p² ) ) = 0

=>  x = -1   or   x = ( p² - r² ) / ( p² - q² )

Answer 2

Answer:

this is wrong question

Step-by-step explanation:

you go man