Question

If a,b are the roots of $x^{2} +px+1 = 0$ and c, d are the roots of $x^{2} +qx+1=0$, then the value of (a-c)(b-c)(a+d)(b+d) is?

1)$p^{2} - q^{2}$
1)$q^{2} - p^{2}$
1)$q^{2} + p^{2}$
4)None of these

Please show the process....!

Answer 1

a, b are the roots of  x² + p x + 1 = 0
 let a = [ -p + √(p² - 4) ]/2      and  b = [-p -√(p²-4)]/2
    So,    a + b = -p  and  a b = 1
     =>  a² + b² = (a+b)² - 2ab = p² - 2

c and d are the roots of x² + q x + 1 = 0
   let c = [-q + √(q²-4) ]/2  and    d = [-q - √(q²-4) ]/2
   also,   c d  = 1  and    c + d = - q
   =>   c² + d² = (c+d)² - 2cd = q² - 2

Now,   (a - c) (b -c) (a+d) (b+d)    :  expand and substitute the above :
      = [ ab + c² -c (a+b) ]  [ ab + d² + d(a+b) ]
      = [ 1 + c² + c p ] [ 1 + d² - p d ]
      = [ 1 + d² - p d + c² + c² d² - c² p d + cp + c p d² - c p² d ]
      = [  1 + c² + d² - p d + (cd)²  - cp *cd + cp + pd *cd - p² cd ]
     = [ 1 + q² - 2  - p d + 1 - p c + c p  + pd - p² ]
     = q² - p²

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well, you could multiply the other terms also: as follows:

   (a-c) (b -c) (a +d ) (b+d) = (a-c)(b+d) (b-c)(a+d)
   = [ab + ad - bc- cd ] [ab + bd - ca - cd ]
   = [ 1 + ad - bc - 1 ] [ 1 + bd - ca - 1]
   = (ad - bc) (bd - ac)
   = abd² - a²cd - b²cd + abc²
   = d² - a² - b² + c²
   = (d² + c²) - (a² + b²)
   =  q² - 2 - (p² - 2)
   = q² - p²