Question

if the roots of x^2+2px+mn=0 are real and equal, show that roots of x^2-2(n+m)x+(m^2+n^2+2p)=0 have real and equal roots

Answer 1

♦ Quadratic Resolution ♦


√√ For real and equal roots , the discriminant 'D' of the equation :->
          ax² + bx + c  = 0 ---> D = b² - 4ac ≥ 0 √

Hence, 
----> D [ x² + 2px + mn ] = [ 2p ]² - 4( mn ) = 4 ( p² - mn ) = 0 √√   [ given ]
                                                                   => p² = mn √√ 
Now, 
---> D [ x^2-2(n+m)x+(m^2+n^2+2p) ]
          = [ 2( m + n ) ]² - 4 [ m² + n² + 2p ] 
          = [ 4m² + 4n² + 8mn - 4m² - 4n² - 8p ]
          = [ 8mn - 8p ]
          = 8 [ p² - p ]       <----- [ mn = p² ]

Hence, D [ x^2-2(n+m)x+(m^2+n^2+2p) ] ≥ 0  --> for all p ≥ 0 √√

Thus, x^2-2(n+m)x+(m^2+n^2+2p) = 0 has real roots for all p ≥ 0 
      ---> And equal roots only at p = { 0 } ∪ { 1 }