# If a,b are the roots of and c, d are the roots of , then the value of (a-c)(b-c)(a+d)(b+d) is? 1) 1) 1) 4)None of these Please show the process....!

1
by ankitkumar0102

2015-04-05T21:13:16+05:30

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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